62nd Ukrainian National Mathematical Olympiad

Suppose such a 2023-gon exists. Let its side be denoted by $a$, so $a^2$ is an integer, and its vertices as $(x_1,y_1)$, $(x_2,y_2)$, ..., $(x_{2023},y_{2023})$. Consider the 2023-gon with the smallest value of $a^2$. We have $(x_i-x_{i+1}{)}^2+(y_i-y_{i+1}{)}^2=a^2$ for each $i$, where $x_{2024}=x_1$, $y_{2024}=y_1$. If $a^2$ is a multiple of 4, then since if the sum of two squares of integers is a multiple of 4, then both numbers are even, we have $x_i\equiv x_{i+1}(\mathrm{mod}2)$, $y_i\equiv y_{i+1}(\mathrm{mod}2)$ for each $i$. But then we can consider a polygon with half the number of vertices with vertices at $(\frac{x_i-x_1}{2},\frac{y_i-y_1}{2})$, whose vertices are also all integer points, and whose side length is $\frac{a}{2}$, obtaining a contradiction.

If $a^2\equiv 2(\mathrm{mod}4)$, then $x_i$ and $x_{i+1}$ have different parity for each $i$, which leads to a contradiction, since we obtain that $x_1$ and $x_1$ have different parity. If $a^2\equiv 1(\mathrm{mod}2)$, then $x_i+y_i$ and $x_{i+1}+y_{i+1}$ have different parity for each $i$, which leads to a contradiction, since we obtain that $x_1+y_1$ and $x_1+y_1$ have different parity.