Suppose $a_1, a_2, a_3, \dots$ is an increasing arithmetic progression of positive integers.  Given that $a_3 = 13$, compute the maximum possible value of \[ a_{a_1} + a_{a_2} + a_{a_3} + a_{a_4} + a_{a_5}. \][i]Proposed by Evan Chen[/i]
1. Let the common difference of the arithmetic progression be \( d \). Given that \( a_3 = 13 \), we can express the first few terms of the sequence as follows:
   \[
   a_1 = 13 - 2d, \quad a_2 = 13 - d, \quad a_3 = 13, \quad a_4 = 13 + d, \quad a_5 = 13 + 2d
   \]

2. The general formula for the \( n \)-th term of the arithmetic progression is:
   \[
   a_n = a_1 + (n-1)d
   \]
   Substituting \( a_1 = 13 - 2d \) into the formula, we get:
   \[
   a_n = (13 - 2d) + (n-1)d = 13 - 2d + nd - d = 13 + (n-3)d
   \]

3. We need to compute \( a_{a_1} + a_{a_2} + a_{a_3} + a_{a_4} + a_{a_5} \). Using the general formula for \( a_n \), we have:
   \[
   a_{a_1} = a_{13-2d}, \quad a_{a_2} = a_{13-d}, \quad a_{a_3} = a_{13}, \quad a_{a_4} = a_{13+d}, \quad a_{a_5} = a_{13+2d}
   \]

4. Substituting these into the general formula:
   \[
   a_{13-2d} = 13 + (13-2d-3)d = 13 + 10d - 2d^2
   \]
   \[
   a_{13-d} = 13 + (13-d-3)d = 13 + 10d - d^2
   \]
   \[
   a_{13} = 13 + (13-3)d = 13 + 10d
   \]
   \[
   a_{13+d} = 13 + (13+d-3)d = 13 + 10d + d^2
   \]
   \[
   a_{13+2d} = 13 + (13+2d-3)d = 13 + 10d + 2d^2
   \]

5. Summing these expressions:
   \[
   a_{a_1} + a_{a_2} + a_{a_3} + a_{a_4} + a_{a_5} = (13 + 10d - 2d^2) + (13 + 10d - d^2) + (13 + 10d) + (13 + 10d + d^2) + (13 + 10d + 2d^2)
   \]
   \[
   = 5 \cdot 13 + 5 \cdot 10d + (-2d^2 - d^2 + d^2 + 2d^2)
   \]
   \[
   = 65 + 50d
   \]

6. To maximize the sum, we need to maximize \( d \). Since \( a_1 \) must be a positive integer, we have:
   \[
   13 - 2d > 0 \implies 2d < 13 \implies d < 6.5
   \]
   Since \( d \) must be an integer, the maximum value of \( d \) is 6.

7. Substituting \( d = 6 \) into the sum:
   \[
   65 + 50 \cdot 6 = 65 + 300 = 365
   \]

The final answer is \( \boxed{365} \).