Okay, so I need to find the sum of all possible values of n given these three equations involving the sequence a1, a2, ..., an. Let me start by writing down the given information:

We have three sums:

1. The sum of the ai's is 96: ∑a_i = 96
2. The sum of the squares of the ai's is 144: ∑a_i² = 144
3. The sum of the cubes of the ai's is 216: ∑a_i³ = 216

And all ai's are positive real numbers. The question asks for the sum of all possible values of n. 

Hmm, okay. Let me recall some inequalities that might be useful here. The Cauchy-Schwarz inequality relates the sum of squares to the square of the sum. Maybe that can help. Also, the Power Mean inequality could be relevant here since we have different powers (1, 2, 3). 

Let me first compute the Cauchy-Schwarz inequality for the sum of squares. The Cauchy-Schwarz inequality states that (∑a_i²)(∑1²) ≥ (∑a_i)². Let's compute that:

Here, ∑a_i² = 144, and ∑1² = n. So, 144 * n ≥ (96)². Let's compute 96². 96*96 is 9216. So, 144n ≥ 9216. Dividing both sides by 144: n ≥ 9216 / 144. Let's compute that. 144*64 = 9216, so 9216/144 = 64. Therefore, n ≥ 64. 

So, the minimum possible value of n is 64. But the question asks for all possible values of n. So, n has to be at least 64. But is there an upper bound? Let's see. Also, considering the third equation: ∑a_i³ = 216. 

Another thought: If all the ai's are equal, then we can compute n. Let's check if that's possible. Suppose all ai = k. Then:

Sum ai = n*k = 96 => k = 96/n

Sum ai² = n*k² = 144 => n*(96/n)² = 144 => n*(9216/n²) = 144 => 9216/n = 144 => n = 9216 / 144 = 64. So, if all ai's are equal, n must be 64, and each ai is 96/64 = 1.5. Then check the sum of cubes: 64*(1.5)^3 = 64*(3.375) = 216. Perfect, that works. So, n=64 is possible. 

But the problem states "the sum of all possible values of n". So, maybe n can be larger? Let's see. If the ai's are not all equal, can we have a larger n? 

Wait, but the Cauchy-Schwarz gave us n ≥ 64, but if n is larger than 64, say 65, would that satisfy the other conditions? Let's check. 

Suppose n=65. Then, using Cauchy-Schwarz, (∑a_i²)(65) ≥ (∑a_i)^2. So, 144*65 ≥ 96². Compute 144*65: 144*60=8640, 144*5=720, total 8640+720=9360. 96² is 9216. So, 9360 ≥ 9216. That's true. So, n=65 is possible? But we need to check if the other sums can also be satisfied. 

But even if C-S doesn't rule out n=65, maybe the other equations (sum of cubes) would impose a stricter condition. Let me think. Let's try to construct a sequence where n=65. 

Suppose we have 64 variables equal to 1.5 (so that their sum is 64*1.5=96), and one variable being 0. Then the sum of squares would be 64*(1.5)^2 + 0 = 64*2.25 = 144. That's good. The sum of cubes would be 64*(1.5)^3 + 0 = 64*3.375 = 216. Perfect. But wait, but the ai's are supposed to be positive real numbers. So, if one of them is zero, that's not positive. So, we can't have a zero. 

Therefore, all ai's must be positive. So, we cannot have a zero. So, n=65 would require that the 65th term is positive, but then the sum of squares would need to be 144. Let's see. If we have 64 terms of 1.5 and one term of x, then sum of squares would be 64*(1.5)^2 + x² = 144. Let's compute 64*(2.25) = 144. So, 144 + x² = 144 => x²=0 => x=0. But again, x must be positive. Therefore, n=65 is impossible. 

Similarly, any n greater than 64 would require that the sum of squares is 144. If we have n=64, then each term is 1.5, sum is 96, sum of squares 144, sum of cubes 216. If we try to have n=65, we need to replace one of the 1.5's with a smaller number and add a new term. Let's say we take one term from 1.5 to 1.5 - ε, and add a new term δ. Then, the sum remains 96, so the decrease in the sum is ε, so the new term must be δ = ε. 

Then, the sum of squares would change by (1.5 - ε)^2 + δ² - 1.5². Let's compute that: (1.5² - 2*1.5*ε + ε²) + ε² - 1.5² = -3ε + ε² + ε² = -3ε + 2ε². The original sum of squares is 144, so the new sum of squares would be 144 -3ε + 2ε². But we need this to equal 144, so -3ε + 2ε² = 0 => ε(2ε - 3) = 0. So, ε=0 or ε=3/2. But ε=3/2 would make the term 1.5 - 3/2 = 0, which is not positive, so ε=0. Therefore, no solution for ε>0, meaning that we can't have n=65 with all positive terms. 

Similarly, for any n>64, the same reasoning applies. Therefore, the only possible value for n is 64. 

Wait, but let me check another approach. Maybe using the Cauchy-Schwarz inequality and the Hölder's inequality. Hölder's inequality relates the sums of different powers. For three sequences, but here we have just one sequence. Wait, Hölder's inequality might not directly help here. Let's think about the relationship between the sums. 

Alternatively, perhaps using the Cauchy-Schwarz in another way. Let me compute the ratio of the sum of cubes to the sum. Hmm, 216 / 96 = 2.25. Wait, that's exactly 1.5. Which is the same as the average of the ai's when they are all 1.5. So, the sum of cubes divided by the sum is equal to the cube of the average. Let's check: (96/64) = 1.5. Then (1.5)^3 = 3.375. Wait, 216 is 64 * 3.375. So, sum of cubes is sum of ai^3 = n*(average)^3. Which is exactly the case when all ai's are equal. 

Therefore, if all ai's are equal, then the sum of cubes is n*(average)^3. Which in this case, n=64, average=1.5, sum of cubes=64*(1.5)^3=216. So, that's the only case when the sum of cubes is exactly n*(average)^3. If the ai's are not all equal, then the sum of cubes would be different. 

But according to the given, sum of cubes is 216, which equals n*(average)^3. Therefore, unless all ai's are equal, the sum of cubes would be different. So, if the sum of cubes is exactly n*(average)^3, does that imply that all ai's are equal? Let me recall that for positive numbers, if the sum of cubes is equal to n*(average)^3, then by the Power Mean inequality, all ai's must be equal. Because the Power Mean inequality states that the mean of the cubes is greater than or equal to the cube of the mean, with equality if and only if all ai's are equal. 

Indeed, the Power Mean inequality says that for positive real numbers, the mean of the k-th powers is greater than or equal to the k-th power of the mean. So, for k=3, we have:

( (a1³ + a2³ + ... + an³)/n )^(1/3) ≥ ( (a1 + a2 + ... + an)/n )

With equality if and only if a1 = a2 = ... = an. 

In our case, (216/n)^(1/3) = (average) = 96/n. Let's check:

(216/n)^(1/3) = (96/n)

Compute both sides:

Left side: (216/n)^(1/3) = (6³/n)^(1/3) = 6 / n^(1/3)

Right side: 96/n

So, 6 / n^(1/3) = 96 / n

Multiply both sides by n^(1/3):

6 = 96 / n * n^(1/3) = 96 * n^(-2/3)

Therefore:

n^(2/3) = 96 / 6 = 16

Therefore:

n = 16^(3/2) = (4²)^(3/2) = 4³ = 64

Therefore, n must be 64. So, that's the only solution. Therefore, the sum of all possible values of n is just 64. 

Wait, but let me verify this once more. If we suppose that the Power Mean inequality holds, then equality occurs only when all ai's are equal. Therefore, since the given sums satisfy the equality case in the Power Mean inequality for k=3, all ai's must be equal, leading to n=64. Therefore, the only possible value of n is 64, so the sum is 64. 

Therefore, the answer is 64.

**Final Answer**
\boxed{64}
Therefore, the sum of all possible values of $n$ is $\boxed{64}$.
</think>