For some $k > 0$ the lines $50x + ky = 1240$ and $ky = 8x + 544$ intersect at right angles at the point $(m,n)$. Find $m + n$.
1. **Determine the slopes of the lines:**
   The given lines are:
   \[
   50x + ky = 1240
   \]
   and
   \[
   ky = 8x + 544
   \]
   To find the slopes, we rewrite each equation in slope-intercept form \(y = mx + b\).

   For the first line:
   \[
   ky = -50x + 1240 \implies y = -\frac{50}{k}x + \frac{1240}{k}
   \]
   The slope \(m_1\) of the first line is:
   \[
   m_1 = -\frac{50}{k}
   \]

   For the second line:
   \[
   ky = 8x + 544 \implies y = \frac{8}{k}x + \frac{544}{k}
   \]
   The slope \(m_2\) of the second line is:
   \[
   m_2 = \frac{8}{k}
   \]

2. **Use the fact that the lines intersect at right angles:**
   Since the lines intersect at right angles, the product of their slopes must be \(-1\):
   \[
   m_1 \cdot m_2 = -1
   \]
   Substituting the slopes:
   \[
   \left(-\frac{50}{k}\right) \left(\frac{8}{k}\right) = -1
   \]
   Simplifying:
   \[
   -\frac{400}{k^2} = -1 \implies \frac{400}{k^2} = 1 \implies k^2 = 400 \implies k = \pm 20
   \]
   Since \(k > 0\), we have:
   \[
   k = 20
   \]

3. **Substitute \(k = 20\) into the original equations:**
   The equations become:
   \[
   50x + 20y = 1240
   \]
   and
   \[
   20y = 8x + 544
   \]

4. **Solve the system of equations:**
   From the second equation:
   \[
   20y = 8x + 544 \implies y = \frac{8}{20}x + \frac{544}{20} \implies y = 0.4x + 27.2
   \]

   Substitute \(y = 0.4x + 27.2\) into the first equation:
   \[
   50x + 20(0.4x + 27.2) = 1240
   \]
   Simplify and solve for \(x\):
   \[
   50x + 8x + 544 = 1240 \implies 58x + 544 = 1240 \implies 58x = 696 \implies x = 12
   \]

5. **Find \(y\) using \(x = 12\):**
   Substitute \(x = 12\) back into \(y = 0.4x + 27.2\):
   \[
   y = 0.4(12) + 27.2 = 4.8 + 27.2 = 32
   \]

6. **Calculate \(m + n\):**
   The point of intersection is \((m, n) = (12, 32)\). Therefore:
   \[
   m + n = 12 + 32 = 44
   \]

The final answer is \(\boxed{44}\)