HMMT November, 2019

Solution:

Suppose $A_{1} A_{2} \ldots A_{n}$ is an equiangular $n$-gon formed from the vertices of a regular 100-gon. Note that the angle $\angle A_{1} A_{2} A_{3}$ is determined only by the number of vertices of the 100-gon between $A_{1}$ and $A_{3}$. Thus in order for $A_{1} A_{2} \ldots A_{n}$ to be equiangular, we require exactly that $A_{1}, A_{3}, \ldots$ are equally spaced and $A_{2}, A_{4}, \ldots$ are equally spaced. If $n$ is odd, then all the vertices must be equally spaced, meaning $n \mid 100$. If $n$ is even, we only need to be able to make a regular $\left(\frac{n}{2}\right)$-gon from the vertices of a 100-gon, which we can do if $n \mid 200$. Thus the possible values of $n$ are $4, 5, 8, 10, 20, 25, 40, 50$, and $100$, for a total of $262$.