Let's determine those positive integers for which $n^{2}-19 n-99$ is a perfect square.
The equation to solve is $n^{2}-19 n-99=k^{2}$, where $k$ and $n$ are positive integers. This can be rewritten as $n^{2}-19 n-\left(99+k^{2}\right)=0$, from which we get $n_{1,2}=\frac{19 \pm \sqrt{4 k^{2}+757}}{2}$. For $n$ to be a positive integer, $4 k^{2}+757$ must be a perfect square: $4 k^{2}+757=l^{2}$, where $l$ is also a positive integer.

$$
757=(l-2 k)(l+2 k)
$$

Since 757 is a prime number, its divisors are: $1, -1, 757, -757$. Given the conditions, $l-2 k < l+2 k$, so there are two possibilities: $l-2 k=-757$ and $l+2 k=-1$ (but this is not possible since $k, l$ are positive integers), or $l-2 k=1$ and $l+2 k=757$.

The solution to the system of equations is $l=379, k=189$. Then $n_{1}=199, n_{2}=-180$. The condition is satisfied by $n=199$ and indeed: $199^{2}-19 \cdot 199-99=35721=189^{2}$.

Eszter Árvai (Szekszárd, Garay J. Gymnasium, 11th grade)