A maximum can be achieved if no three diagonals meet at the same point. In this case, any intersection point has a bijective relation with four distinct vertices. This is because assuming there are four vertices $A$, $B$, $C$, and $D$ in that order but not necessarily next to each other, then diagonals $AC$ and $BD$ will create an intersection point. On the other hand, any intersection point created by diagonals $AC$ and $BD$ must have their ending points selected from four distinct vertices of the $n$-polygon.

Then, it follows that the answer is $\boxed{\binom{n}{4}}$.