Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz=1,$ $x+\frac{1}{z}=5,$ and $y+\frac{1}{x}=29.$ Then $z+\frac{1}{y}=\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
1. Given the equations:
   \[
   xyz = 1, \quad x + \frac{1}{z} = 5, \quad y + \frac{1}{x} = 29
   \]

2. From the second equation, solve for \(x\):
   \[
   x + \frac{1}{z} = 5 \implies x = 5 - \frac{1}{z}
   \]

3. From the third equation, solve for \(y\):
   \[
   y + \frac{1}{x} = 29 \implies y = 29 - \frac{1}{x}
   \]

4. Substitute \(x = 5 - \frac{1}{z}\) into the equation \(xyz = 1\):
   \[
   \left(5 - \frac{1}{z}\right) y z = 1
   \]

5. Simplify the equation:
   \[
   (5z - 1) y = 1 \implies y = \frac{1}{5z - 1}
   \]

6. Substitute \(y = \frac{1}{5z - 1}\) into \(y + \frac{1}{x} = 29\):
   \[
   \frac{1}{5z - 1} + \frac{1}{5 - \frac{1}{z}} = 29
   \]

7. Simplify \(\frac{1}{5 - \frac{1}{z}}\):
   \[
   \frac{1}{5 - \frac{1}{z}} = \frac{z}{5z - 1}
   \]

8. Substitute back into the equation:
   \[
   \frac{1}{5z - 1} + \frac{z}{5z - 1} = 29 \implies \frac{1 + z}{5z - 1} = 29
   \]

9. Solve for \(z\):
   \[
   1 + z = 29(5z - 1) \implies 1 + z = 145z - 29 \implies 1 + 29 = 145z - z \implies 30 = 144z \implies z = \frac{30}{144} = \frac{5}{24}
   \]

10. Substitute \(z = \frac{5}{24}\) back into the equation \(z + \frac{1}{y}\):
    \[
    y = \frac{1}{5z - 1} = \frac{1}{5 \cdot \frac{5}{24} - 1} = \frac{1}{\frac{25}{24} - 1} = \frac{1}{\frac{25}{24} - \frac{24}{24}} = \frac{1}{\frac{1}{24}} = 24
    \]

11. Calculate \(z + \frac{1}{y}\):
    \[
    z + \frac{1}{y} = \frac{5}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4}
    \]

12. Therefore, \(z + \frac{1}{y} = \frac{1}{4}\), where \(m = 1\) and \(n = 4\). Thus, \(m + n = 1 + 4 = 5\).

The final answer is \(\boxed{5}\)