1012 red points?

i think that the good bound is the seeling of 2022/2 to prove this you can use induction to find a more general case n/2

The answer is $1012$. To get the bound, simply consider a triangulation of red points. For the construction, simply do $n=5$ (which is pretty much forced) and generalise. In particular we take a regular $n$ gon $A_1A_2\cdots A_n$ and for each $A_k$ ($k\ge 3$) we put a blue point in triangle $A_1A_2A_k$, in the region nearest to $A_k$. A bit of casework shows that this works.