1. Use conservation of mechanical energy

Since both bodies roll without slipping, their initial total kinetic energy converts completely into gravitational potential energy at the highest point.

So, $mgh = K_{\text{trans}} + K_{\text{rot}}$

with $K_{\text{trans}} = \frac{1}{2}mv^2, \qquad K_{\text{rot}} = \frac{1}{2}I\omega^2$

Also, for rolling without slipping, $\omega = \frac{v}{R}$

2. For the solid sphere

Moment of inertia of a solid sphere about its center: $I_{\text{sphere}} = \frac{2}{5}mR^2$

Thus, $K_{\text{rot}} = \frac{1}{2}\cdot \frac{2}{5}mR^2 \cdot \frac{v^2}{R^2} = \frac{1}{5}mv^2$

So total initial kinetic energy is $K_{\text{sphere}} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$

At maximum height $h_1$, $mgh_1 = \frac{7}{10}mv^2$

Hence, $h_1 = \frac{7v^2}{10g}$

3. For the hollow cylinder

For a hollow cylinder (ring), $I_{\text{cyl}} = mR^2$

Thus, $K_{\text{rot}} = \frac{1}{2}mR^2 \cdot \frac{v^2}{R^2} = \frac{1}{2}mv^2$

So total initial kinetic energy is $K_{\text{cyl}} = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$

At maximum height $h_2$, $mgh_2 = mv^2$

Hence, $h_2 = \frac{v^2}{g}$

4. Find the ratio

$h_1:h_2 = \frac{7v^2}{10g} : \frac{v^2}{g} = \frac{7}{10}:1 = 7:10$

Given, $h_1:h_2 = \frac{n}{10}$

So, $\frac{n}{10} = \frac{7}{10} \Rightarrow n=7$

5. Final answer

$\boxed{7}$