Junior Balkan Mathematical Olympiad

Suppose there are $r$ red points among some $n$ points. Since any pair of them can lie on at most two blue-centered unit circles, it means that the number $b$ of blue points can be at most $2\left(\genfrac{}{}{0ex}{}{r}{2}\right)=r(r-1)$. Since $b+r=n$, this leads to condition $r+r(r-1)=r^2\ge n$, i.e. $r\ge \lceil \sqrt{n}\rceil$, so $b\le n-\lceil \sqrt{n}\rceil$.

A simple model is given by $r=\lceil \sqrt{n}\rceil$ red points of coordinates $R_i(r_i,0)$, with $0<r_i<2$, for all $i=1,2,\dots ,r$. Take $n-r$ blue points among those $r(r-1)$ given by all pairs $(i,j)$, $1\le i<j\le r$, and of coordinates $B_{i,j}(x_{i,j},b_{i,j})$ and $B_{i,j}^{\prime }(x_{i,j},-b_{i,j})$, with $$x_{i,j}=\frac{r_i+r_j}{2}\text{and}{b}_{i,j}=\sqrt{1-{\left(\frac{r_i-r_j}{2}\right)}^2}.$$ It is trivial to check that all blue points $B_{i,j}$, $B_{i,j}^{\prime }$ are distinct, while on unit circles of these centers lie points $R_i$ and $R_j$, and only them.

The choices made warrant that $n-r\le r(r-1)$, since it comes to $n\le \lceil \sqrt{n}{\rceil }^2$, and that $(r_i-r_j{)}^2/4<1$. For the given $n=2009$, the answer is thus that the largest possible number of blue points is $2009-\lceil \sqrt{2009}\rceil =2009-45=1964$.