Let $a$ and $b$ be positive integers that satisfy $ab-7a-11b+13=0$. What is the minimum possible value of $a+b$?
1. Start with the given equation:
   \[
   ab - 7a - 11b + 13 = 0
   \]

2. Use Simon's Favorite Factoring Trick (SFFT) to factor the equation. First, rearrange the terms to facilitate factoring:
   \[
   ab - 7a - 11b + 13 = 0 \implies ab - 7a - 11b = -13
   \]

3. Add and subtract 77 to complete the factorization:
   \[
   ab - 7a - 11b + 77 - 77 = -13 \implies ab - 7a - 11b + 77 = 64
   \]

4. Factor the left-hand side:
   \[
   (a - 11)(b - 7) = 64
   \]

5. We need to find the factor pairs of 64 and determine the pair that gives the minimum sum for \(a + b\). The factor pairs of 64 are:
   \[
   (1, 64), (2, 32), (4, 16), (8, 8), (16, 4), (32, 2), (64, 1)
   \]

6. Calculate \(a\) and \(b\) for each factor pair:
   \[
   \begin{aligned}
   &a - 11 = 1 \implies a = 12, \quad b - 7 = 64 \implies b = 71 \quad \Rightarrow a + b = 12 + 71 = 83 \\
   &a - 11 = 2 \implies a = 13, \quad b - 7 = 32 \implies b = 39 \quad \Rightarrow a + b = 13 + 39 = 52 \\
   &a - 11 = 4 \implies a = 15, \quad b - 7 = 16 \implies b = 23 \quad \Rightarrow a + b = 15 + 23 = 38 \\
   &a - 11 = 8 \implies a = 19, \quad b - 7 = 8 \implies b = 15 \quad \Rightarrow a + b = 19 + 15 = 34 \\
   &a - 11 = 16 \implies a = 27, \quad b - 7 = 4 \implies b = 11 \quad \Rightarrow a + b = 27 + 11 = 38 \\
   &a - 11 = 32 \implies a = 43, \quad b - 7 = 2 \implies b = 9 \quad \Rightarrow a + b = 43 + 9 = 52 \\
   &a - 11 = 64 \implies a = 75, \quad b - 7 = 1 \implies b = 8 \quad \Rightarrow a + b = 75 + 8 = 83 \\
   \end{aligned}
   \]

7. The minimum value of \(a + b\) from the above calculations is 34.

The final answer is \(\boxed{34}\).