Since all diagonals from one vertex divide an $n$-gon into $n-2$ disjoint triangles, at least $n-2$ points are necessary.

We claim that it is possible to mark $n-2$ points so that the given condition is satisfied. Denote the vertices of the given $n$-gon with $A_1, A_2, \dots, A_n$. Draw all the diagonals of the given $n$-gon and color the areas bounded by the diagonals $A_1A_k$, $A_kA_n$ and $A_{k-1}A_{k+1}$ for each $k \in \{2, 3, \dots, n-1\}$.

If we mark one point in each colored area then every triangle with vertices at vertices of the $n$-gon will contain at least one of the marked points. Indeed, triangle $A_lA_kA_m$ contains the whole colored area bounded by diagonals $A_lA_k$, $A_kA_n$ and $A_{k-1}A_{k+1}$ for every $1 \le l < k < m \le n$. Hence, the triangle also contains the corresponding marked point.