Actually,there are $\frac{3}{2}n(2017-n)$ isosceles triangles which have different colors of vertices.

As number of all isosceles triangles are fixed, we will prove instead that the number of non-monochromatic isosceles triangles does not depend on coloring. We will show that if we change $2017$ to any $m\in\mathbb{Z}_{>0}$ such that $\gcd(m,6)=1$ then the number of non-monochromatic triangles are $\frac 32 n(m-n)$. Label each vertex as $1,2,\hdots, m$ in counterclockwise order. We see that three vertices $a,b,c$ will form the isosceles iff $x\equiv\frac{y+z}{2}\pmod m$ for some permutation $(x,y,z)$ of $(a,b,c)$. Let $S$ be a set of sets $\{a,b,c\}$ such that vertices $a,b,c$ form an isosceles triangle, and not all of them have the same color. We use double counting with the number of pairs $(T,e)$ where $T\in S$ and $e$ is an edge that joins vertices with different colors of $T$. For the first way, we see that for any $T\in S$, we can choose exactly $2$ choices of $e$. Thus, the number is equal to $2|T|$. For the second way, we we choose $e=\{a,b\}$ from any pairs of different color vertices first, which has $n(m-n)$ choices. (There are $n$ red vertices and $m-n$ blue vertices.) Then, we choose $c$ such that $\{a,b,c\}\in S$. The only possible choices of $c$ are $2a-b,\frac{a+b}{2},2b-a\pmod m$. Observe that because $a\not\equiv b\pmod m$ and $6\nmid m$, these three numbers are different in modulo $m$. Thus, there are exactly $3$ choices to choose $T$ for each $e$, so the number of pairs $(T,e)$ is $3n(m-n)$. Therefore, $2|S|=3n(m-n)$, yielding that the number of non-monochromatic triangles are exactly $\frac 32n(m-n)$, independent from the way of vertices coloring.