Integers $1, 2, 3, ... ,n$, where $n > 2$, are written on a board. Two numbers $m, k$ such that $1 < m < n, 1 < k < n$ are removed and the average of the remaining numbers is found to be $17$. What is the maximum sum of the two removed numbers?
1. Let's denote the sum of the first \( n \) integers as \( S \). The sum \( S \) is given by the formula:
   \[
   S = \frac{n(n+1)}{2}
   \]
2. When two numbers \( m \) and \( k \) are removed, the sum of the remaining numbers is:
   \[
   S - m - k = \frac{n(n+1)}{2} - m - k
   \]
3. The average of the remaining \( n-2 \) numbers is given to be 17. Therefore, we can write:
   \[
   \frac{\frac{n(n+1)}{2} - m - k}{n-2} = 17
   \]
4. Multiplying both sides by \( n-2 \) to clear the fraction, we get:
   \[
   \frac{n(n+1)}{2} - m - k = 17(n-2)
   \]
5. Simplifying the right-hand side:
   \[
   \frac{n(n+1)}{2} - m - k = 17n - 34
   \]
6. Multiplying the entire equation by 2 to eliminate the fraction:
   \[
   n(n+1) - 2m - 2k = 34n - 68
   \]
7. Rearranging the equation to isolate \( m + k \):
   \[
   n^2 + n - 34n + 68 = 2m + 2k
   \]
   \[
   n^2 - 33n + 68 = 2m + 2k
   \]
   \[
   2m + 2k = n^2 - 33n + 68
   \]
   \[
   m + k = \frac{n^2 - 33n + 68}{2}
   \]
8. We need to find the maximum value of \( m + k \). Since \( 2 \leq m, k \leq n-1 \), it follows that:
   \[
   4 \leq m + k \leq 2n - 2
   \]
9. Therefore, we must have:
   \[
   \frac{n^2 - 33n + 68}{2} \leq 2n - 2
   \]
10. Multiplying both sides by 2 to clear the fraction:
    \[
    n^2 - 33n + 68 \leq 4n - 4
    \]
11. Rearranging the inequality:
    \[
    n^2 - 37n + 72 \leq 0
    \]
12. Solving the quadratic inequality \( n^2 - 37n + 72 \leq 0 \) using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
    \[
    n = \frac{37 \pm \sqrt{37^2 - 4 \cdot 1 \cdot 72}}{2 \cdot 1}
    \]
    \[
    n = \frac{37 \pm \sqrt{1369 - 288}}{2}
    \]
    \[
    n = \frac{37 \pm \sqrt{1081}}{2}
    \]
    \[
    n = \frac{37 \pm 32.86}{2}
    \]
    \[
    n = \frac{37 + 32.86}{2} \approx 34.93 \quad \text{and} \quad n = \frac{37 - 32.86}{2} \approx 2.07
    \]
13. Since \( n \) must be an integer and \( n > 2 \), the maximum value of \( n \) is 34.
14. Substituting \( n = 34 \) back into the expression for \( m + k \):
    \[
    m + k = \frac{34^2 - 33 \cdot 34 + 68}{2}
    \]
    \[
    m + k = \frac{1156 - 1122 + 68}{2}
    \]
    \[
    m + k = \frac{102}{2}
    \]
    \[
    m + k = 51
    \]

The final answer is \(\boxed{51}\).