Obviously, there are exactly $n-1$ different nontrivial rotations of a regular $n$-gon (rotations of $\frac{2\pi i}{n}$ angles, for $1 \le i \le n-1$). Denote the number of blue and red vertices of the polygon by $b$ and $r$, respectively $(a+b=n)$, and the number of vertices that have different colors before and after the rotation of $\frac{2\pi j}{n}$ angle by $t_j$. According to the problem statement, $t_j < \frac{32}{100}n$ for $j = 1, 2, \dots, n-1$. On the other hand, during the $n-1$ rotations each blue vertex coincides with each red vertex exactly once and conversely, each red vertex coincides with each blue vertex exactly once. Therefore, from one hand the total number of pair of points with opposite colors after all of the nontrivial rotations is $\sum_{j=1}^{n-1} t_j$, and $2rb = 2r(n-r)$ from another. Hence,
$$2r(n-r) = \sum_{j=1}^{n-1} t_j < (n-1) \times \frac{32}{100}n < \frac{32}{100}n^2,$$
or equivalently,
$$r^2 - nr + \frac{16}{100}n^2 > 0 \Leftrightarrow \left(r - \frac{n}{2}\right)^2 > \frac{9}{100}n^2 \Leftrightarrow \left|r - \frac{n}{2}\right| > \frac{3}{10}n,$$
which completes the proof.