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 $2\binom{r}{2} = r(r-1)$. 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 $B'_{i,j}(x_{i,j}, -b_{i,j})$, with
$$x_{i,j} = \frac{r_i + r_j}{2} \quad \text{and} \quad b_{i,j} = \sqrt{1 - \left(\frac{r_i - r_j}{2}\right)^2}.$$
It is trivial to check that all blue points $B_{i,j}$, $B'_{i,j}$ 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 $2009 - \lceil \sqrt{2009} \rceil = 2009 - 45 = 1964$.