Answer: for all $n$ that give a remainder of 2 when divided by 5.

Solution. If a board $n \times n$ can be cut into one $2 \times 2$ square and some number of strips of five cells, then $n^{2}=2^{2}+5 m$, from which $n$ gives a remainder of 2 or 3 when divided by 5. Suppose that $n=5 k+3$ and the board can be cut in the required way. Rotate it so that the square adjoins the top side of the board. Write ones in the cells of the top row, twos in the cells of the next row, and so on. Note that the sum of the numbers in five consecutive rows is divisible by 5, since

$$
n i+n(i+1)+n(i+2)+n(i+3)+n(i+4)=5 n(i+2) \vdots 5
$$

Therefore, the remainder of the division by 5 of the sum of all the numbers written is

$$
(n+2 n+3 n) \bmod 5=6(5 k+3) \bmod 5=3
$$

On the other hand, in each strip the sum of the numbers is divisible by five, and in the square the sum of the numbers is $1+1+2+2=6$. Thus, the remainder of the division by 5 of the sum of all the numbers written is 1, and we get a contradiction.

If $n=5 k+2$, then we can cut out a corner $2 \times 2$ square, cut the top strip $2 \times 5 k$ into horizontal strips of five cells, and cut the rectangle $5 k \times(5 k+2)$ into vertical strips of five cells.
5. For which $n$ can a square grid $n \times n$ be divided into one $2 \times 2$ square and some number of strips of five cells, such that the square is adjacent to the side of the board?