1. For which natural numbers $n$ can a square grid $n \times n$ be cut into rectangles $1 \times 1$ and $2 \times 3$ such that there are an equal number of rectangles of each size?
Answer: all $n>7$, multiples of 7.

Solution. Let the board can be cut into $m$ squares $1 \times 1$ and $m$ rectangles $2 \times 3$. Then $n^{2}=m+6 m=7 m$, from which $n^{2} \vdots 7$ and $n \vdots 7$, and $m=\frac{n^{2}}{7}$. Therefore, $n$ that are not divisible by 7 do not fit.

Let $n \vdots$ 7. Notice that for $n \geqslant 14$ the board $n \times n$ can be cut into squares of size $14 \times 14$ and $21 \times 21$. Therefore, it is enough to check that $n=14$ and $n=21$ satisfy the condition of the problem, while $n=7$ does not. For such $n$, the maximum number of non-overlapping rectangles of size $2 \times 3$ that can fit on the board $n \times n$ is 28, 70, and 6, while 28, 63, and 7 are required, respectively. Thus, the case $n=7$ does not fit. For $n=14$ and $n=21$, 28 and 63 rectangles of size $2 \times 3$ can be cut out, and the remaining part of the board can be cut into single cells. Therefore, $n=14$ and $n=21$ satisfy the condition.