Okay, let's see. The problem is asking for the probability that the reciprocal of a randomly selected positive odd integer less than 2010 gives a terminating decimal. Then, we need to express this probability as a reduced fraction a/b and find a + b. Hmm, alright. Let me break this down step by step.

First, I remember that a fraction in its simplest form will have a terminating decimal if and only if the denominator (after simplifying) has no prime factors other than 2 or 5. So, for the reciprocal of an integer, say n, 1/n will terminate if and only if n's prime factors are only 2s and/or 5s. But wait, the problem specifies that the integer is a positive odd integer less than 2010. Since it's odd, it can't have 2 as a factor. Therefore, for 1/n to terminate, n must be a power of 5. Because if n is an odd integer and after simplifying 1/n has a denominator with only 2s and 5s, but since n is odd, the denominator can't have any 2s. Therefore, n must be a power of 5. So, the task reduces to finding all odd integers less than 2010 that are powers of 5, then dividing that count by the total number of positive odd integers less than 2010. Then, simplify the fraction and add the numerator and denominator.

Alright, let's formalize this. Let me denote the set of positive odd integers less than 2010 as S. The size of S is the number of positive odd integers less than 2010. The number of such integers is equal to floor((2010 - 1)/2) + 1. Wait, let me check. The positive odd integers start at 1, 3, 5, ..., up to the largest odd integer less than 2010. Since 2010 is even, the largest odd integer less than 2010 is 2009. The number of terms in this sequence is (2009 - 1)/2 + 1 = 2008/2 + 1 = 1004 + 1 = 1005. So, there are 1005 positive odd integers less than 2010. Therefore, the denominator b is 1005, but we need to check if 1005 and the numerator a are coprime. Wait, no, the probability is the number of favorable outcomes over total outcomes, so the numerator is the number of odd integers less than 2010 that are powers of 5, and the denominator is 1005. Then, we reduce the fraction a/b where a is the count of such integers and b is 1005. Then, a + b is the answer.

So, first, let's find all odd integers less than 2010 that are powers of 5. Let's list the powers of 5:

5^0 = 1
5^1 = 5
5^2 = 25
5^3 = 125
5^4 = 625
5^5 = 3125

Wait, 5^5 is 3125, which is greater than 2010. Therefore, the powers of 5 that are odd and less than 2010 are 1, 5, 25, 125, 625. So, that's 5 numbers. Therefore, the numerator a is 5. So, the probability is 5/1005. Now, we need to simplify this fraction. Let's see, 5 and 1005. 1005 divided by 5 is 201. So, 5/1005 simplifies to 1/201. Therefore, a is 1 and b is 201, so a + b is 202. Wait, but hold on. Let me verify again.

Wait, 5^0 is 1, which is included. 5^1 is 5, 5^2 is 25, 5^3 is 125, 5^4 is 625, 5^5 is 3125. 3125 is over 2010, so stop at 5^4. So that's 1, 5, 25, 125, 625. That's 5 numbers. So, 5 favorable outcomes. Total outcomes are 1005. So, 5/1005. Divide numerator and denominator by 5: 1/201. So, a=1, b=201. Then a + b = 202. Therefore, the answer is 202.

Wait, but let me check once again. Let's verify the count of odd integers less than 2010. The first odd integer is 1, then 3, 5, ..., up to 2009. The number of terms is (2009 - 1)/2 + 1 = 2008/2 + 1 = 1004 + 1 = 1005. That's correct. So total is 1005. Then, powers of 5: 5^0=1, 5^1=5, 5^2=25, 5^3=125, 5^4=625. 5^5=3125 which is over 2010, so 5 numbers. So 5/1005 simplifies to 1/201. Therefore, 1 + 201 = 202. Seems correct.

But let me check if there's any mistake here. For example, is 1 considered a power of 5? Yes, 5^0 is 1. So 1 is included. And 1 is an odd integer, so yes. Then 5, 25, 125, 625. All these numbers are less than 2010. 625 is 625, 5^4 is 625, 5^5 is 3125, which is over. So yes, 5 numbers. So, I think that's correct.

Alternatively, maybe the problem is considering only numbers that are only divisible by 2 and 5? But no, since the integers are odd, they can't be divisible by 2. Therefore, the only prime factor allowed is 5. Hence, only powers of 5. So, the count is indeed 5. Therefore, the answer is 202.

Wait, but let me think again. The problem says "the reciprocal of a randomly selected positive odd integer less than 2010 gives a terminating decimal." So, the reciprocal 1/n is terminating. So, n must be a product of 2s and 5s. But since n is odd, it can't have 2s. Therefore, n must be a power of 5. So, yes, n must be 1, 5, 25, 125, 625. So, 5 numbers. Therefore, 5/1005 reduces to 1/201. So, a + b = 1 + 201 = 202. Therefore, the answer is 202.

But let me cross-verify once more. Let's take some examples. For example, n=1: 1/1=1.0, which is terminating. n=5: 1/5=0.2, terminating. n=25: 1/25=0.04, terminating. n=125: 1/125=0.008, terminating. n=625: 1/625=0.0016, terminating. All of these are terminating. What about n=25*5=125, which is already included. So, all these numbers are indeed powers of 5. So, 5 numbers. So, the count is correct.

Total numbers: 1005. So, probability 5/1005 = 1/201. Therefore, a=1, b=201. So, a + b=202. Therefore, the answer is 202. I think that's solid. I can't see any mistakes here. Let me check if 1005 is indeed the correct count.

The positive odd integers less than 2010: the first term is 1, the last term is 2009. The number of terms in the arithmetic sequence starting at 1, ending at 2009, with a common difference of 2 is ((2009 - 1)/2) + 1 = (2008/2) + 1 = 1004 + 1 = 1005. Correct. So, denominator is 1005. So, 5/1005 simplifies by dividing numerator and denominator by 5: 5 ÷ 5 =1, 1005 ÷5=201. So, 1/201. Therefore, a + b=1 + 201=202. So, the answer is 202. I think that's correct.
Therefore, the probability that the reciprocal of a randomly selected positive odd integer less than 2010 gives a terminating decimal is $\frac{1}{201}$. Therefore, $a + b = 1 + 201 = 202$.

\boxed{202}
</think>