32. (9th grade) Find $2 n+1$ consecutive integers that have the property that the sum of the squares of the first $\boldsymbol{n}+1$ numbers is equal to the sum of the squares of the next $n$ numbers.
32. Let $N$ be the first number we are looking for. Using the condition of the problem, we can write:

$$
N^{2}+(N+1)^{2}+(N+2)^{2}+\ldots+(N+n)^{2}=(N+n+1)^{2}+\ldots+(N+2 n)^{2}
$$

$N^{2}=2 n^{3}+2 n^{2} N+n^{2}$, from which $N=n^{2} \pm\left(n^{2}+n\right)$.

If $N_{1}=2 n^{2}+n$, then for $n=1, N_{1}=3$ we get the numbers 3, 4, and 5. For $n=2, N_{1}=10$ we get five numbers: $10, 11, 12, 13, 14$, and so on.

If we take the second root $N_{2}=-n$, then for $n=1, N_{2}=-1$, we get three numbers: $-1, 0, +1$. For $n=2, N_{2}=-2$, we get five numbers: $-2, -1, 0, +1, +2$, and so on.