First, notice that none of the triangles formed by taking three vertices of the given polygon is equilateral. If there were one, we would have the same number of vertices of the original polygon between any two of the triangle's vertices. This would imply that $3$ divides $2014$, which is false.

Let us now prove that each segment (whose endpoints are vertices of the given polygon) belongs to exactly $3$ different isosceles triangles.

Consider an arbitrary segment. Its endpoints divide the circumcircle of the polygon into two arcs containing $2015$ vertices. One of the arcs contains an odd number, and the other contains an even number of vertices. Therefore, there exists a unique isosceles triangle with the chosen segment as its base (the third vertex is the midpoint of the arc containing an odd number of vertices). In a similar fashion, we infer that the chosen segment is a base in exactly two isosceles triangles – there are two possible choices of the third segment, both on the longer arc.

Let $B$ be the number of blue points, so that $B + R = 2017$. Let $d_b, d_r, d_s$, be the number of blue (both endpoints blue), red, and two-colour segments, respectively. Furthermore, let $t_b, t_r, t_{sb}, t_{sr}$ be the number of isosceles triangles with the vertices coloured all blue, all red, two blue – one red, and two red – one blue, respectively.

From the fact that each segment belongs to three isosceles triangles, we get:
$$\begin{aligned} 3d_b &= 3t_b + t_{sb}, \\ 3d_r &= 3t_r + t_{sr}, \\ 3d_s &= 2t_{sb} + 2t_{sr}. \end{aligned}$$
This implies $3d_b + 3d_r - \frac{3}{2}d_s = 3t_b + 3t_r$, i.e.
$$t_b + t_r = d_b + d_r - \frac{1}{2}d_s = \frac{1}{2} \cdot (B(B-1) + R(R-1) - BR),$$
which depends only on the number of the red and blue points, but not on their arrangement.