### Component 1
<description>
Understanding the problem and setting the initial goal.
</description>

<plan>
Okay, let me try to figure out this problem. The question is asking for the largest positive integer with only even digits that is less than 10,000 and is a multiple of 9. Hmm, let's break this down step by step.
</plan>

<prompt>
Understand the problem statement and set the initial goal for solving it.
</prompt>

<execution>
First, since the number has to be less than 10,000, it can be up to four digits. So we're looking for a four-digit number, right? But all digits must be even. The even digits are 0, 2, 4, 6, and 8. So each digit in the number must be one of these. Got that.

Next, it needs to be a multiple of 9. From what I remember, a number is divisible by 9 if the sum of its digits is also divisible by 9. So whatever number we come up with, the sum of its digits must be a multiple of 9. That seems like a crucial point here.

Alright, so the goal is to find the largest four-digit number with all even digits where the sum of the digits is a multiple of 9.
</execution>


### Component 2
<description>
Checking if the largest possible number with even digits (8888) is a multiple of 9.
</description>

<plan>
Let's start by thinking about the largest possible four-digit number with even digits. The largest even digit is 8, so the largest four-digit number with even digits would be 8888.
</plan>

<prompt>
Check if 8888 is a multiple of 9 by calculating the sum of its digits.
</prompt>

<execution>
But is 8888 a multiple of 9? Let's check the sum: 8 + 8 + 8 + 8 = 32. 32 divided by 9 is not an integer, so 32 isn't a multiple of 9. So 8888 doesn't work.
</execution>


### Component 3
<description>
Attempting to adjust the digits of 8888 to make it a multiple of 9.
</description>

<plan>
Okay, so we need to find the next largest number with even digits. Since we want the largest possible number, we should try to keep the leftmost digits as large as possible, and adjust the rightmost digits if needed.
</plan>

<prompt>
Try to adjust the digits of 8888 to make the sum of its digits a multiple of 9.
</prompt>

<execution>
Let's start from 8888 and work our way down.

The sum of the digits needs to be a multiple of 9. Let's see what the closest multiple of 9 is to 32. The multiples of 9 around 32 are 27 (which is 9*3) and 36 (9*4). 32 is between 27 and 36. But 36 is higher than 32, so we can't reach that. The next possible sum is 27. So the sum of digits needs to be 27. Wait, but 27 is 9*3, which is a multiple of 9. So if we can adjust the digits of 8888 to make their sum 27 instead of 32, then that number would be a multiple of 9.
</execution>


### Component 4
<description>
Realizing the impossibility of reducing the sum by an odd number.
</description>

<plan>
How much do we need to reduce the sum by? 32 - 27 = 5.
</plan>

<prompt>
Calculate how much the sum needs to be reduced and determine if it's possible.
</prompt>

<execution>
So we need to decrease the sum of digits by 5. But we need to do this by changing the digits from 8 to smaller even digits (0, 2, 4, 6). Each time we reduce a digit by some even amount, we reduce the sum by that amount. For example, changing an 8 to a 6 reduces the sum by 2; changing an 8 to a 4 reduces the sum by 4, etc.

We need a total reduction of 5. But here's the problem: all digits must be even, and any change we make has to reduce the sum by an even number, right? Because we're replacing an even digit with another even digit. So replacing 8 (even) with 6 (even) subtracts 2 (even), replacing with 4 subtracts 4 (even), and so on. So all the reductions will be even numbers. But 5 is an odd number. Wait, that's a problem. If we need to reduce the total sum by 5, which is odd, but every individual digit change reduces the sum by an even number, then the total reduction would have to be even as well. Therefore, it's impossible to reduce 32 by 5 to get 27 because 5 is odd. Therefore, 27 is not achievable.
</execution>


### Component 5
<description>
Identifying the correct target sum for the digits.
</description>

<plan>
Then what's the next possible multiple of 9 below 27? That would be 18.
</plan>

<prompt>
Determine the next possible multiple of 9 below 32 and check if it can be achieved with even digits.
</prompt>

<execution>
Wait, but 18 is way lower. Wait, 27 is the next lower multiple of 9 from 32. Wait, but 27 is 9*3, which is 27. Then 32 - 27 = 5, which is impossible. Then is there no multiple of 9 between 18 and 32? Wait, 9*2=18, 9*3=27, 9*4=36. So between 18 and 36, we have 27. But we can't reach 27 from 32 because we need to subtract 5, which is odd. So maybe 36 is too high, 27 is unreachable. Therefore, does that mean there is no four-digit number with even digits whose digits sum to 9, 18, or 27? Wait, 18 is 9*2, which is lower. So maybe 18 is possible? Let's check.

But if we can't reach 27, maybe we need to go lower. Let's think again. Wait, perhaps we need to check if there is a possible sum that is a multiple of 9 and can be achieved by even digits. Let me think. The maximum sum with even digits for a four-digit number is 8+8+8+8=32. The minimum is 0+0+0+0=0, but since it's a positive integer, the minimum is probably 2 (like 2000). Wait, but even 2000 has a digit sum of 2. So possible sums are even numbers from 0 up to 32. Wait, but 0 isn't possible because it's a positive integer. So the possible sums are even numbers from 2 to 32. Now, which of these are multiples of 9? Let's list the multiples of 9 in that range: 9, 18, 27. But 9, 18, 27 are all odd or even? 9 is odd, 18 is even, 27 is odd. So in the range of even sums (since all digits are even, the sum must be even), the only possible multiple of 9 is 18. Because 18 is even, and 9 and 27 are odd. Therefore, the sum must be 18. That's the key. So even though 27 is a multiple of 9, it's odd, so the sum of even digits can't be odd. Therefore, the only possible sum that is a multiple of 9 and even is 18.
</execution>


### Component 6
<description>
Finding the largest number with a digit sum of 18.
</description>

<plan>
So that changes things. So the target digit sum is 18.
</plan>

<prompt>
Find the largest four-digit number with even digits where the sum of the digits is 18.
</prompt>

<execution>
So we need to find the largest four-digit number with even digits where the digits sum to 18. Okay, that's a different problem. So instead of trying to reduce 32 by 14 (to get 18), which seems like a lot, but maybe there is a higher number than just going all the way down. Let's see.

Our task now is to find the largest four-digit number with all even digits (0,2,4,6,8) where the sum of digits is 18. Since we need the largest possible number, we want the leftmost digits to be as large as possible. So starting with the thousands place, we want it to be 8 if possible, then 6, etc. , but we have to adjust the remaining digits so that the total sum is 18.

Let me approach this systematically. The thousands digit should be as large as possible. Let's start with 8.

Case 1: Thousands digit is 8. Then the remaining three digits (hundreds, tens, units) must sum to 18 - 8 = 10. These digits must be even (0,2,4,6,8). We need the hundreds digit to be as large as possible, then tens, then units.

So thousands digit is 8. Then hundreds digit should be as large as possible. Let's try 8 again. If hundreds digit is 8, then the remaining two digits must sum to 10 - 8 = 2. The tens digit should be as large as possible. The largest even digit for tens place is 2, but 2 is too much? Wait, 2 is allowed. Wait, remaining sum is 2. So tens digit can be 2, then units digit 0. So the number would be 8 (thousands), 8 (hundreds), 2 (tens), 0 (units) = 8820. The sum is 8+8+2+0=18. That works. But is there a larger number?

Wait, let's see. If hundreds digit is 8, tens digit can be 0 and units digit 2. So 8802, but that's smaller than 8820. So 8820 is better. Alternatively, if hundreds digit is 6, then remaining sum is 10 - 6 = 4. Then tens digit can be 4, units 0. So 8640. The sum is 8+6+4+0=18. 8640 is less than 8820. If hundreds digit is 4, remaining sum is 10 - 4 = 6. Then tens digit 6, units 0: 8460. That's smaller. Similarly, hundreds digit 2: 10 -2=8, tens digit 8, units 0: 8280. Still less than 8820. Hundreds digit 0: 10-0=10, but we need two digits summing to 10. Largest possible is 8 and 2, so 80 (tens and units), so number would be 8082. Which is smaller. So 8820 is the largest in the case where thousands digit is 8 and hundreds digit is 8.

But wait, maybe there's a way to have a higher number by adjusting the digits differently. Let's check. If thousands digit is 8, hundreds digit is 8, tens digit is 2, units digit is 0. 8820. If we try to increase the tens digit, but we can't because 8+8+something would exceed the remaining sum. Wait, the remaining sum after thousands and hundreds digits (8+8=16) is 18-16=2. So tens and units digits have to sum to 2. So the maximum tens digit is 2, then units 0. So 8820 is indeed the maximum in this scenario.
</execution>


### Component 7
<description>
Exploring other possibilities with different thousands digits.
</description>

<plan>
But let's check other possibilities where thousands digit is 8, but hundreds digit is less than 8. For example, thousands digit 8, hundreds digit 6.
</plan>

<prompt>
Explore numbers with thousands digits less than 8 to see if a larger number can be found.
</prompt>

<execution>
Then remaining sum is 18 - 8 -6 = 4. So tens and units digits need to sum to 4. The largest possible tens digit here is 4, then units 0. So the number is 8640. Alternatively, tens digit 2, units digit 2: 8622. Which is smaller than 8640. So 8640 is better.

Similarly, thousands digit 8, hundreds digit 4: remaining sum 6. Largest tens digit 6, units 0: 8460. Thousands digit 8, hundreds digit 2: remaining sum 8. Tens digit 8, units 0: 8280. Thousands digit 8, hundreds digit 0: remaining sum 10. Tens and units digits sum to 10. Largest possible is 8 and 2, so 8082. So all these numbers are smaller than 8820.

Therefore, with thousands digit 8, the maximum is 8820.

But wait, maybe if we reduce the thousands digit to 6, we can get a larger number overall? Wait, no. 8820 is starting with 8, which is larger than any number starting with 6. So 8820 is larger. So maybe 8820 is the answer. But let's check other cases to be thorough.

Case 2: Thousands digit is 6. Then the remaining three digits must sum to 18 -6 =12. Again, we want the hundreds digit to be as large as possible. Let's try 8. So thousands digit 6, hundreds digit 8: remaining sum 12 -8=4. Then tens digit 4, units 0: 6840. Sum is 6+8+4+0=18. Alternatively, tens digit 2, units 2: 6822. Which is smaller. So 6840 is better.

If hundreds digit is 6: remaining sum 12-6=6. Tens digit 6, units 0: 6660. Sum 6+6+6+0=18. If hundreds digit is 4: remaining sum 12-4=8. Tens digit 8, units 0: 6480. Sum 6+4+8+0=18. If hundreds digit is 2: remaining sum 12-2=10. Tens and units digits sum to 10. Largest tens digit 8, units 2: 6282. If hundreds digit is 0: remaining sum 12. Tens and units digits sum to 12. The maximum would be 8 and 4: 6084. So numbers here are 6840, 6660, 6480, 6282, 6084. The largest is 6840. But 6840 is less than 8820, so 8820 is still larger.
</execution>


### Component 8
<description>
Continuing exploration with different thousands digits.
</description>

<plan>
Case 3: Thousands digit is 4. Then remaining three digits must sum to 18-4=14.
</plan>

<prompt>
Continue exploring numbers with thousands digits less than 8 to find a larger number.
</prompt>

<execution>
Hundreds digit as large as possible, 8. Then remaining sum 14-8=6. Tens digit 6, units 0: 4860. Sum 4+8+6+0=18. If hundreds digit 6: remaining sum 14-6=8. Tens digit 8, units 0: 4680. Hundreds digit 4: remaining sum 14-4=10. Tens digit 8, units 2: 4482. Hundreds digit 2: remaining sum 14-2=12. Tens digit 8, units 4: 4284. Hundreds digit 0: remaining sum 14. Tens digit 8, units 6: 4086. All these numbers are smaller than 8820.

Case 4: Thousands digit is 2. Then remaining digits sum to 18-2=16. Hundreds digit 8: remaining sum 16-8=8. Tens digit 8, units 0: 2880. Sum 2+8+8+0=18. Hundreds digit 6: remaining sum 16-6=10. Tens digit 8, units 2: 2682. Hundreds digit 4: remaining sum 16-4=12. Tens digit 8, units 4: 2484. Hundreds digit 2: remaining sum 16-2=14. Tens digit 8, units 6: 2286. Hundreds digit 0: remaining sum 16. Tens digit 8, units 8: 2088. All of these are smaller than 8820.
</execution>


### Component 9
<description>
Validating the case where the thousands digit is 0.
</description>

<plan>
Case 5: Thousands digit is 0. But since it's a four-digit number, thousands digit can't be 0.
</plan>

<prompt>
Validate if a four-digit number can have a thousands digit of 0.
</prompt>

<execution>
So this case is invalid.
</execution>


### Component 10
<description>
Confirming 8820 as the largest number and verifying its properties.
</description>

<plan>
Therefore, the largest number we found so far is 8820. But wait, let's confirm that there isn't a larger number.
</plan>

<prompt>
Confirm that 8820 is the largest number and verify its properties.
</prompt>

<execution>
For example, is there a number between 8820 and 8888 with even digits and digit sum 18? Let's check.

The next number after 8820 would be 8822, but digits must be even. 8822 has digits 8,8,2,2. Sum is 8+8+2+2=20, which is not 18. 8824: digits 8,8,2,4. Sum is 22. Not 18. 8826: 24. 8828: 26. 8830: invalid because 3 is odd. Similarly, 8840: digits 8,8,4,0. Sum is 20. 8860: sum is 22. 8880: sum is 24. So none of these have a digit sum of 18. What if we decrease the thousands digit? Wait, 8820 is the highest possible with thousands digit 8, hundreds digit 8. If we try to adjust the other digits. For example, 8802: sum is 18, but 8802 is smaller than 8820. Similarly, 8820 is larger. What about 8802 vs 8820? 8820 is higher.

Wait, another thought. Maybe instead of having two 8s, we can have three 8s? Let's see. If thousands, hundreds, and tens digits are 8, then units digit would need to be 18 -8-8-8= -6, which is impossible. So that's not possible. So three 8s can't work. What about two 8s? 8,8, something, something. We had 8,8,2,0 which gives 18. If we try 8,6,4,0: sum is 18. But 8640 is less than 8820. So 8820 is still higher.

Wait, another approach. Let's think of the digits as variables: thousands (T), hundreds (H), tens (Te), units (U). All even digits. T+H+Te+U=18. We need the number T H Te U to be as large as possible. The largest number is achieved when T is maximized, then H, then Te, then U. So we start with T=8. Then H is maximized, given T=8 and remaining sum 10. Then H=8, leaving sum 2 for Te and U. Then Te=2, U=0. So 8820. If H=6, then remaining sum is 4. So Te=4, U=0: 8640. So 8820 is the largest possible. If we set T=8, H=8, Te=2, U=0, that's the maximum.

Alternatively, is there a way to have T=8, H=8, Te=0, U=2? That would be 8802. But 8820 is larger than 8802. So no. Similarly, any other combination would result in a smaller number.

But let's check all possible combinations with T=8 and H=8. The remaining two digits need to sum to 2. The possible pairs (Te, U) are (2,0) and (0,2). So 8820 and 8802. The larger one is 8820.

Therefore, 8820 seems to be the candidate.
</execution>


### Component 11
<description>
Verifying that 8820 meets all the problem's criteria.
</description>

<plan>
Let's verify the sum: 8+8+2+0=18. 18 divided by 9 is 2.
</plan>

<prompt>
Verify that 8820 meets all the problem's criteria.
</prompt>

<execution>
So 8820 is indeed a multiple of 9. And all digits are even. And it's a four-digit number less than 10,000. So that seems to fit all the requirements.
</execution>


### Component 12
<description>
Thoroughly checking for any larger numbers.
</description>

<plan>
But just to make sure there are no larger numbers. For example, maybe if we decrease a higher place value and increase a lower one.
</plan>

<prompt>
Thoroughly check for any larger numbers that meet the criteria.
</prompt>

<execution>
For example, decreasing thousands digit by 2 (to 6) and increasing another digit. But 6 in thousands place would make the number start with 6, which is less than 8, so the entire number would be smaller. So that's not helpful.

Alternatively, maybe if we have more 8s in the lower digits. Let's see. Suppose thousands digit is 8, hundreds digit is 6, tens digit is 4, units digit is 0. 8640. Sum is 8+6+4+0=18. That's valid. But 8640 is less than 8820.

Alternatively, thousands digit 8, hundreds digit 4, tens digit 6, units digit 0. 8460. Still less.

Another thought: what if the digits are 8, 8, 8, but then the fourth digit is -6? No, that's invalid. So no.

Alternatively, 8, 8, 1, 1. But digits must be even. So that's not allowed.

Wait, another idea: maybe the sum can be 18 by having different combinations. Let's see. For example, 8,6,6,8. Sum is 8+6+6+8=28. That's too much. 8,6,4,0=18. 8,8,2,0=18. 6,6,6,0=18. 6,6,4,2=18. Let's see if there's a higher number than 8820 with sum 18.

For example, 8682: digits 8,6,8,2. Sum is 8+6+8+2=24. Not 18. 8680: sum 22. 8640: sum 18. 8622: sum 18. 8484: sum 24. 8448: sum 24. 8406: sum 18. But 8406 is lower than 8820.

Alternatively, 8280: sum 18. But 8280 is lower than 8820.

Wait, is there a number like 8, 2, 8, 0? 8280. Sum 18. But 8280 is lower than 8820. So no.

Alternatively, 8, 4, 6, 0: 8460. Sum 18. Also lower.

Hmm. It seems that 8820 is indeed the largest possible. But just to be thorough, let's check if 8820 is the maximum. Let's consider numbers starting with 88. The next digit is 2, then 0. 8820. If we try to make the third digit larger, like 4, but then the sum would be 8+8+4=20, so the units digit would need to be -2, which is invalid. So we can't have a larger third digit. Similarly, if we try to make the third digit 8, the sum would be 8+8+8=24, which would require the units digit to be -6, impossible. So no.

Alternatively, if we lower the hundreds digit from 8 to 6, then the remaining digits can be higher. For example, thousands digit 8, hundreds digit 6: remaining sum 4. So tens digit 4, units 0. 8640. But 8640 is less than 8820.

Alternatively, if thousands digit 8, hundreds digit 8, tens digit 2, units digit 0. That's 8820. If we can make tens digit 2 into 6 by adjusting other digits, but how? Let's see. To increase the tens digit from 2 to 6, we need to add 4. That would require subtracting 4 from another digit. But the thousands and hundreds digits are already 8, which is the maximum. If we subtract 4 from the hundreds digit (8 to 4), then add 4 to the tens digit (2 to 6). Then the number becomes 8460. Which is smaller. Similarly, if we subtract from thousands digit, making it 4, and add to tens digit: 4860. Even smaller. So no gain there.

Therefore, 8820 is the largest possible number meeting the criteria.

But wait, let's check another possibility. What if the thousands digit is 8, the hundreds digit is 6, the tens digit is 4, and the units digit is 0: 8640. Sum is 18. If we can rearrange the digits to get a higher number. For example, swapping hundreds and tens digits: 8460. That's lower. Not helpful. Or swapping hundreds and units: 8604. But 8604 is higher than 8640? Wait, 8604 vs 8640. 8604 is lower because the hundreds digit is 6 vs 4. Wait, 8640 is 8 thousand 6 hundred 40, whereas 8604 is 8 thousand 6 hundred 04. So 8640 is higher. So no improvement.

Alternatively, making the number 8, 6, 6, 8. But digits must be even. 8668. Sum is 8+6+6+8=28. Not 18.

Alternatively, 8, 6, 2, 2: 8622. Sum 18. But 8622 is lower than 8820.

Another idea: Let's consider numbers where the thousands digit is 8, hundreds digit is 8, and the last two digits are 2 and 0. So 8820 is one, and 8802 is another. Since 8820 is higher, that's better. If we can't make the last two digits higher without exceeding the sum, then that's the max.

Alternatively, if the last two digits are both 1s, but they have to be even. So 0,2,4, etc.

Wait, another approach: Let's consider the maximum number with even digits that is less than 10,000. That's 8888. Since that's not a multiple of 9, we need the next largest number with even digits. But how do we find that?

Instead of starting from 8888 and subtracting 2 each time (since numbers must be even), but checking each one for digit sum being a multiple of 9. But this could take a long time. Instead, perhaps using the digit sum approach is better.

But according to our earlier analysis, the maximum possible number is 8820, and that seems to check out. Let's verify again.

8820:

Digits: 8,8,2,0. All even. Sum: 8+8+2+0=18. 18 is divisible by 9. Therefore, 8820 is a multiple of 9. And it's a four-digit number with all even digits. And there's no larger number than 8820 that meets these criteria.
</execution>


### Component 13
<description>
Final confirmation that 8820 is the solution.
</description>

<plan>
So I think that's the answer.

But just to be absolutely sure, let's check the next few numbers after 8820 to see if any of them meet the criteria.
</plan>

<prompt>
Confirm that 8820 is the solution by checking the next few numbers.
</prompt>

<execution>
Next number after 8820 with even digits: 8822. Check digits: 8,8,2,2. Sum is 20. Not divisible by 9. 8824: sum 22. 8826: sum 24. 8828: sum 26. 8830: has a 3, which is odd. Invalid. 8840: sum 20. 8842: sum 22. 8844: sum 24. 8846: sum 26. 8848: sum 28. 8860: sum 22. 8862: sum 24. 8864: sum 26. 8866: sum 28. 8868: sum 30. 8880: sum 24. 8882: sum 26. 8884: sum 28. 8886: sum 30. 8888: sum 32. All these numbers either have a digit sum that isn't a multiple of 9 or contain odd digits. So none of them work.

Therefore, 8820 is indeed the next number after 8888 that meets the criteria. Therefore, the answer should be 8820.
</execution>


### Component 14
<description>
Presenting the final answer.
</description>

<plan>
None
</plan>

<prompt>
Present the final answer.
</prompt>

<execution>
**Final Answer**
The largest positive integer with only even digits that is less than $10,000$ and is a multiple of $9$ is \boxed{8820}.
</execution>
