99. Answer: $n=9$.

Let $a_{n}$ be the side length, and $D_{n}$ and $d_{n}$ be the lengths of the largest and smallest diagonals of a regular $n$-gon. For $n=4$ and $n=5$, all diagonals are equal. For $n=6$ and $n=7$, $D_{n}-a_{n}=2 A K=D_{8}-d_{8}$. For $n=9$ (Fig. 55,6), similarly, we get $\angle A B K=30^{\circ}$, so $A B=a_{9}=2 A K=D_{9}-d_{9}$.

We will further consider the $n$-gon inscribed in a circle of radius 1. For $n>9$, obviously, $D_{n} \geqslant D_{9}, d_{n}D_{9}-d_{9}=a_{9}>a_{n}$.
99. In a regular $n$-gon ( $n>5$ ), the difference between the largest and smallest diagonals is equal to the side. Find $n$.