Extremely classical. Not a competition topic. Answer is $4$. Nice Hadwiger proof for $n>4$ ($n\neq 6$) not to be.

can u explain.

This is IMO Long List 1986-P8.

Assume there exists an $n$-gon all of whose vertices are lattice points. Work in the complex plane; label the vertices counterclockwise by $p_1,...,p_n$. The center of the $n$-gon, call it $q$, is the centroid of the vertices: $q=\frac{p_1+...+p_n}{n}$ Thus all the vertices and the center of the pentagon have both rational imaginary and real components. Let $\omega^n=1;\omega\neq1$. Then $p_2=q+\omega(p_1-q) \to \omega=\frac{p_2-q}{p_1-q}$ The RHS has both rational imaginary and real components. But $\omega=\cos(\frac{2\pi}{n})+i\sin(\frac{2\pi}{n})$ Both $\cos(\frac{2\pi}{n}), \sin(\frac{2\pi}{n})$ are rational only if $n=4$, by Niven's Theorem.