Given that $O$ is a regular octahedron, that $C$ is the cube whose vertices are the centers of the faces of $O$, and that the ratio of the volume of $O$ to that of $C$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime integers, find $m+n$.
1. **Set the side length of the octahedron to 1 for simplicity.** 
   - The volume of a regular octahedron with side length \( a \) is given by:
     \[
     V_O = \frac{\sqrt{2}}{3} a^3
     \]
   - For \( a = 1 \):
     \[
     V_O = \frac{\sqrt{2}}{3}
     \]

2. **Determine the height of the octahedron.**
   - An octahedron can be divided into two pyramids with a square base.
   - The height of each pyramid can be found using a 45-45-90 triangle:
     \[
     \text{Height} = \frac{\sqrt{2}}{2}
     \]

3. **Find the side length of the cube.**
   - The vertices of the cube are the centers of the faces of the octahedron.
   - The center of each face of the octahedron is the circumcenter of an equilateral triangle.
   - The height of an equilateral triangle with side length 1 is:
     \[
     \text{Height} = \frac{\sqrt{3}}{2}
     \]
   - The circumcenter is \(\frac{1}{3}\) of the way up the height of the triangle:
     \[
     \text{Distance from base to circumcenter} = \frac{\sqrt{3}}{6}
     \]
   - The height of the pyramid above the circumcenter is:
     \[
     \text{Height above circumcenter} = \frac{\sqrt{3}}{3}
     \]

4. **Calculate the side length of the cube.**
   - The side length of the cube is the distance between two adjacent face centers of the octahedron.
   - This distance is the hypotenuse of a right triangle with legs \(\frac{\sqrt{2}}{3}\) and \(\frac{\sqrt{3}}{3}\):
     \[
     \text{Side length of cube} = \sqrt{\left(\frac{\sqrt{2}}{3}\right)^2 + \left(\frac{\sqrt{3}}{3}\right)^2} = \sqrt{\frac{2}{9} + \frac{3}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}
     \]

5. **Calculate the volume of the cube.**
   - The volume of a cube with side length \( s \) is \( s^3 \):
     \[
     V_C = \left(\frac{\sqrt{5}}{3}\right)^3 = \frac{5\sqrt{5}}{27}
     \]

6. **Find the ratio of the volumes.**
   - The ratio of the volume of the octahedron to the volume of the cube is:
     \[
     \text{Ratio} = \frac{V_O}{V_C} = \frac{\frac{\sqrt{2}}{3}}{\frac{5\sqrt{5}}{27}} = \frac{\sqrt{2}}{3} \cdot \frac{27}{5\sqrt{5}} = \frac{27\sqrt{2}}{15\sqrt{5}} = \frac{27\sqrt{2}}{15\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{27\sqrt{10}}{75} = \frac{9\sqrt{10}}{25}
     \]

7. **Simplify the ratio and find \( m+n \).**
   - The simplified ratio is:
     \[
     \frac{9}{2}
     \]
   - Therefore, \( m = 9 \) and \( n = 2 \):
     \[
     m+n = 9+2 = 11
     \]

The final answer is \( \boxed{11} \).