Principal Skinner is thinking of two integers $m$ and $n$ and bets Superintendent Chalmers that he will not be able to determine these integers with a single piece of information.  Chalmers asks Skinner the numerical value of $mn+13m+13n-m^2-n^2$.  From the value of this expression alone, he miraculously determines both $m$ and $n$.  What is the value of the above expression.
1. We start with the given expression:
   \[
   mn + 13m + 13n - m^2 - n^2
   \]
   Let this expression be equal to \( Z \):
   \[
   mn + 13m + 13n - m^2 - n^2 = Z
   \]

2. Rearrange the terms to group the quadratic and linear terms:
   \[
   m^2 + n^2 - mn - 13m - 13n = -Z
   \]

3. Multiply the entire equation by 2 to facilitate completing the square:
   \[
   2m^2 + 2n^2 - 2mn - 26m - 26n = -2Z
   \]

4. Rewrite the left-hand side by completing the square:
   \[
   (m-n)^2 + (m-13)^2 + (n-13)^2 = 338 - 2Z
   \]
   This step uses the identity for completing the square:
   \[
   (m-n)^2 + (m-13)^2 + (n-13)^2 = m^2 + n^2 - 2mn + m^2 - 26m + 169 + n^2 - 26n + 169
   \]
   Simplifying the right-hand side:
   \[
   2m^2 + 2n^2 - 2mn - 26m - 26n + 338 = 338 - 2Z
   \]

5. For the expression to have a unique solution, the right-hand side must be zero:
   \[
   338 - 2Z = 0
   \]

6. Solving for \( Z \):
   \[
   2Z = 338
   \]
   \[
   Z = 169
   \]

The final answer is \(\boxed{169}\).