The summation $\sum_{k=1}^{360} \frac{1}{k \sqrt{k+1} + (k+1)\sqrt{k}}$ is the ratio of two relatively prime positive integers $m$ and $n$. Find $m + n$.
1. We start with the given summation:
   \[
   \sum_{k=1}^{360} \frac{1}{k \sqrt{k+1} + (k+1)\sqrt{k}}
   \]

2. To simplify the expression, we rationalize the denominator. Multiply the numerator and the denominator by the conjugate of the denominator:
   \[
   \frac{1}{k \sqrt{k+1} + (k+1)\sqrt{k}} \cdot \frac{k \sqrt{k+1} - (k+1)\sqrt{k}}{k \sqrt{k+1} - (k+1)\sqrt{k}}
   \]

3. The numerator becomes:
   \[
   k \sqrt{k+1} - (k+1)\sqrt{k}
   \]

4. The denominator becomes:
   \[
   (k \sqrt{k+1} + (k+1)\sqrt{k})(k \sqrt{k+1} - (k+1)\sqrt{k}) = (k \sqrt{k+1})^2 - ((k+1)\sqrt{k})^2
   \]

5. Simplify the denominator:
   \[
   (k^2 (k+1)) - ((k+1)^2 k) = k^3 + k^2 - (k^3 + 2k^2 + k) = k^2 - k^2 - k = -k
   \]

6. Thus, the expression simplifies to:
   \[
   \frac{k \sqrt{k+1} - (k+1)\sqrt{k}}{-k} = \frac{(k \sqrt{k+1} - (k+1)\sqrt{k})}{k} \cdot \frac{1}{-1} = \frac{\sqrt{k+1}}{k} - \frac{\sqrt{k}}{k+1}
   \]

7. Rewrite the summation:
   \[
   \sum_{k=1}^{360} \left( \frac{\sqrt{k+1}}{k} - \frac{\sqrt{k}}{k+1} \right)
   \]

8. Notice that this is a telescoping series. When expanded, most terms will cancel out:
   \[
   \left( \frac{\sqrt{2}}{1} - \frac{\sqrt{1}}{2} \right) + \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{3} \right) + \cdots + \left( \frac{\sqrt{361}}{360} - \frac{\sqrt{360}}{361} \right)
   \]

9. The series simplifies to:
   \[
   \left( \frac{\sqrt{2}}{1} - \frac{\sqrt{1}}{2} \right) + \left( \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{3} \right) + \cdots + \left( \frac{\sqrt{361}}{360} - \frac{\sqrt{360}}{361} \right)
   \]

10. All intermediate terms cancel, leaving:
    \[
    1 - \frac{1}{19}
    \]

11. Simplify the final expression:
    \[
    1 - \frac{1}{19} = \frac{19}{19} - \frac{1}{19} = \frac{18}{19}
    \]

12. The fraction $\frac{18}{19}$ is already in its simplest form, where $m = 18$ and $n = 19$. Therefore, $m + n = 18 + 19 = 37$.

The final answer is $\boxed{37}$