Let's determine the sum of the following numbers:

$$
-1^{2}, 2^{2},-3^{2}, 4^{2}, \ldots-(2 n-1)^{2},(2 n)^{2}
$$
The given numbers are grouped as follows:

$$
\left(2^{2}-1^{2}\right),\left(4^{2}-3^{2}\right), \ldots\left(\left(2 n^{2}\right)-(2 n-1)^{2}\right)
$$

or

$$
(2-1)(2+1),(4-3)(4+3), \ldots(2 n-2 n+1)(2 n+2 n-1)
$$

or

$$
3,7, \ldots(4 n-1)
$$

We have obtained an arithmetic progression, the sum of which is:

$$
S=\frac{n}{2}(4 n+2)=n(2 n+1)
$$

(Zsófia Nendtvich, Budapest.)

Number of solutions: 30.