Let $A_1, A_2, \dots, A_{2016}$ be $2016$ points on the circle which are coloured red and the remaining blue. Let $n \ge 3$ and let $B_1, B_2, \dots, B_n$ be a regular $n$-sided polygon inscribed in this circle with the vertices chosen in anti-clock-wise direction. We place $B_1$ at $A_1$. (It is possible, in this position, some other $B$'s also coincide with some other $A$'s.) Rotate the polygon in anti-clock-wise direction gradually till some $B$'s coincide with (an equal number of) $A$'s second time. We again rotate the polygon in the same direction till some $B$'s coincide with an equal number of $A$'s third time, and so on until we return to the original position, i.e., $B_1$ at $A_1$. We see that the number of rotations will not be more than $2016 \times n$, that is, at most these many times some $B$'s would have coincided with an equal number of $A$'s. Since the interval $(0, 360^\circ)$ has infinitely many points, we can find a value $\alpha^\circ \in (0, 360^\circ)$ through which the polygon can be rotated from its initial position such that no $B$ coincides with any $A$. This gives a $n$-sided regular polygon having only blue vertices.

Alternate Solution:

Consider a regular $2017 \times n$-gon on the circle; say, $A_1A_2A_3\cdots A_{2017n}$. For each $j$, $1 \le j \le 2017$, consider the points $\{A_k : k \equiv j \pmod{2017}\}$. These are the vertices of a regular $n$-gon, say $S_j$. We get $2017$ regular $n$-gons; $S_1, S_2, \dots, S_{2017}$. Since there are only $2016$ red points, by pigeon-hole principle there must be some $n$-gon among these $2017$ which does not contain any red point. But then it is a blue $n$-gon.