If we double the number of sides in a polygon, then the difference between the diagonals and sides increases by 99. How many sides does the polygon have?
The number of diagonals in an $n$-sided polygon is $\frac{n(n-3)}{2}$, and thus for a $2n$-sided polygon, it is $n(2n-3)$. According to the problem,

$$
\frac{n(n-3)}{2}-n+99=n(2n-3)-2n
$$

which simplifies to

$$
n^{2}-5n+198=4n^{2}-10n
$$

Rearranging gives

$$
3n^{2}-5n-198=0
$$

from which

$$
n_{1}=9, \quad\left[n_{2}=-\frac{22}{3}\right]
$$

$n_{2}$ cannot be considered due to the nature of the problem, so only the 9-sided polygon satisfies the requirements of the problem. Jármai Erzsébet (Bp. XIII., Ép-gép. techn. I. o. t.)