Let's determine all positive integers $n$ for which

$$
1^{2}+2^{2}+\ldots+n^{2}=1+2+\ldots+(2 n-1)+2 n
$$
On the left side, the sum of the squares of numbers from 1 to $n$ stands, which can be written in the following closed form:

$$
\frac{n(n+1)(2 n+1)}{6}
$$

On the right side, however, the sum of integers from 1 to $2 n$ stands, which is $\frac{2 n(2 n+1)}{2}=n(2 n+1)$.

Using these, we get the following equation:

$$
\frac{n(n+1)(2 n+1)}{6}=n(2 n+1)
$$

From which, by rearranging and simplifying with $n(2 n+1) \neq 0$: $n+1=6, n=5$.

Indeed,

$$
1+4+9+16+25=55=1+2+3+\cdots+9+10
$$

The condition of the problem is satisfied only by the positive integer $n=5$.

Pach Péter Pál (Fazekas M. Fóv. Gyak. Gimn., 8. o.t.)