Volume $A$ equals one fourth of the sum of the volumes $B$ and $C$, while volume $B$ equals one sixth of the sum of the volumes $A$ and $C$. There are relatively prime positive integers $m$ and $n$ so that the ratio of volume $C$ to the sum of the other two volumes is $\frac{m}{n}$. Find $m+n$.
1. We start with the given equations:
   \[
   A = \frac{B + C}{4}
   \]
   \[
   B = \frac{A + C}{6}
   \]

2. Substitute \( B \) from the second equation into the first equation:
   \[
   A = \frac{\frac{A + C}{6} + C}{4}
   \]

3. Simplify the right-hand side:
   \[
   A = \frac{\frac{A + C + 6C}{6}}{4} = \frac{\frac{A + 7C}{6}}{4} = \frac{A + 7C}{24}
   \]

4. Multiply both sides by 24 to clear the fraction:
   \[
   24A = A + 7C
   \]

5. Rearrange to isolate \( A \):
   \[
   24A - A = 7C \implies 23A = 7C \implies A = \frac{7C}{23}
   \]

6. Now substitute \( A \) back into the second equation:
   \[
   B = \frac{\frac{7C}{23} + C}{6}
   \]

7. Simplify the right-hand side:
   \[
   B = \frac{\frac{7C + 23C}{23}}{6} = \frac{\frac{30C}{23}}{6} = \frac{30C}{138} = \frac{5C}{23}
   \]

8. We now have expressions for \( A \) and \( B \) in terms of \( C \):
   \[
   A = \frac{7C}{23}, \quad B = \frac{5C}{23}
   \]

9. The problem asks for the ratio of volume \( C \) to the sum of the other two volumes:
   \[
   \text{Ratio} = \frac{C}{A + B} = \frac{C}{\frac{7C}{23} + \frac{5C}{23}} = \frac{C}{\frac{12C}{23}} = \frac{C \cdot 23}{12C} = \frac{23}{12}
   \]

10. The ratio is \(\frac{23}{12}\), and since \(m\) and \(n\) are relatively prime, we have \(m = 23\) and \(n = 12\).

11. Therefore, \(m + n = 23 + 12 = 35\).

The final answer is \(\boxed{35}\)