Let $a$ and $b$ be integer solutions to $17a+6b=13$. What is the smallest possible positive value for $a-b$?
1. We start with the given equation:
   \[
   17a + 6b = 13
   \]
   We need to find integer solutions for \(a\) and \(b\).

2. Express \(a\) in terms of \(b\):
   \[
   a = \frac{13 - 6b}{17}
   \]
   For \(a\) to be an integer, the numerator \(13 - 6b\) must be divisible by 17. Therefore, we need:
   \[
   13 - 6b \equiv 0 \pmod{17}
   \]
   Simplifying, we get:
   \[
   6b \equiv 13 \pmod{17}
   \]

3. To solve for \(b\), we need the multiplicative inverse of 6 modulo 17. We find the inverse \(x\) such that:
   \[
   6x \equiv 1 \pmod{17}
   \]
   Using the Extended Euclidean Algorithm, we find that the inverse of 6 modulo 17 is 3 because:
   \[
   6 \cdot 3 = 18 \equiv 1 \pmod{17}
   \]

4. Multiply both sides of the congruence \(6b \equiv 13 \pmod{17}\) by 3:
   \[
   b \equiv 13 \cdot 3 \pmod{17}
   \]
   \[
   b \equiv 39 \pmod{17}
   \]
   Simplifying, we get:
   \[
   b \equiv 5 \pmod{17}
   \]
   Therefore, \(b = 5 + 17k\) for some integer \(k\).

5. Substitute \(b = 5 + 17k\) back into the expression for \(a\):
   \[
   a = \frac{13 - 6(5 + 17k)}{17}
   \]
   Simplify the numerator:
   \[
   a = \frac{13 - 30 - 102k}{17}
   \]
   \[
   a = \frac{-17 - 102k}{17}
   \]
   \[
   a = -1 - 6k
   \]

6. We need to find the smallest positive value for \(a - b\):
   \[
   a - b = (-1 - 6k) - (5 + 17k)
   \]
   \[
   a - b = -1 - 6k - 5 - 17k
   \]
   \[
   a - b = -6 - 23k
   \]

7. To find the smallest positive value, we need \(-6 - 23k > 0\):
   \[
   -6 - 23k > 0
   \]
   \[
   -23k > 6
   \]
   \[
   k < -\frac{6}{23}
   \]
   Since \(k\) must be an integer, the smallest integer \(k\) that satisfies this inequality is \(k = -1\).

8. Substitute \(k = -1\) into \(a - b\):
   \[
   a - b = -6 - 23(-1)
   \]
   \[
   a - b = -6 + 23
   \]
   \[
   a - b = 17
   \]

The final answer is \(\boxed{17}\).