Bariz - Problem

The answer is negative. Let $\sqrt{s}$ be the common sidelength. If $s$ is odd, then color lattice points like a checkerboard. Then neighboring vertices of the polygon are opposite colors, but $2023$ — odd. If $s \equiv 2 \pmod{4}$ color the rows alternating black and white; then neighboring vertices must be opposite colors again, since $s = \Delta x^2 + \Delta y^2 \implies \Delta x, \Delta y \equiv 1 \pmod{4}$. If $s \equiv 0 \pmod{4}$ then $\Delta x, \Delta y$ are even, so we can appropriately shift and scale the polygon by $\tfrac{1}{2}$ to get another such polygon and repeat until we reach the other cases.