Obviously $200n\leq \frac {n(n-3)}2$ so $n\geq 403$. More generally, we prove that if $n>150$ and if no diagonal intersects more than $10000$ others, we have drawn at most $200(n-51)$ diagonals. Proof If $n<350$ we have $\frac {n(n-3)}2\leq 200(n-51)$ and it's obvious. Now if $n>350$ and we have drawn more than $200(n-51)$ diagonals then we have a vertex from which exit at least $350$ diagonals. Take the $200$th diagonal. Now it breaks the polygon into two smaller ones. Let the polygons have sides $k,l$($k+l=n+2)$. We note $k,l>150$ Then by induction hypothesis the number of diagonals in first polygon is at most $200(k-51)$ and in second at most $200(l-51)$. Summing since we have at least $200(n-51)$ diagonals we get at least $10000$ diagonals intorsecting the $200th$ diagonal contradiction QED

iura wrote: If $n<350$ we have $\frac {n(n-3)}2\leq 200(n-51)$ and it's obvious. For $n=349$ it is wrong. But I think that this solution may be corrected, though we did not know such solution.

Well, we can replace that $n=348$ and do a little bit corrections. But I think this doesn't matter much..

What's the official solution?

I think, it is quite incorrect question on the forum, which is devoted to problem solving...