Selection tests for the Gulf Mathematical Olympiad, 2013

Let $P_{1} P_{2} \cdots P_{n}$ be a convex cyclic non-regular polygon with the measures of all its internal angles equal and let $O$ be its circumcenter. Because all the angles are equal, all the arcs $\widehat{P_{i} P_{i+2}}$, for $i=1, \ldots, n$, have the same length. Hence, $\angle P_{i} O P_{i+2} = \frac{4\pi}{n}$, for all $i=1, \ldots, n$, since the polygon is convex.

Let $\theta = \angle P_{1} O P_{2}$. We have
$$\angle P_{2i-1} O P_{2i} = \angle P_{1} O P_{2} + \angle P_{2} O P_{2i} - \angle P_{1} O P_{2i-1} = \angle P_{1} O P_{2} = \theta.$$
If $n$ is an odd integer,
$$\theta = \angle P_{n} O P_{1} = \frac{n+1}{2} \angle P_{n} O P_{2} = \frac{n+1}{2} \cdot \frac{4\pi}{n} = \frac{2\pi}{n} = \angle P_{1} O P_{2} \quad \bmod 2\pi.$$
This means that the polygon is regular, which contradicts the hypothesis.

If $n$ is even, any value of $\theta$ with $0 < \theta < \frac{4\pi}{n}$ and $\theta \neq \frac{2\pi}{n}$ defines a unique non-regular such polygon.

Therefore, the possible values of $n$ are all even positive integers $n \geq 4$.