Considering the $6n$ evenly spaces points on the circle, call a point exterior if it is one of the chosen points which splits the circle and interior if it is in the interior of one of the arcs. Let the dual splitting of the circle be the one generated by changing every interior point to exterior and vice versa. If no points are diametrically opposite, the dual overlaps the original splitting exactly if placed diametrically opposite; i.e., the splitting is self-dual. If we list the splitting by listing the arc lengths in order (e.g., $1,3,2,1,2,3$), some thought shows that $1$s and $3$s must alternate with arbitrary numbers of $2$s placed inside, and the dual is formed by changing every $1$ to a $3$ and vice versa. Since a $1$ must be diametrically opposite a $3$, $n$ must be odd; but since the same number of $2$s must appear between $1$s and $3$s as between $3$s and $1$s, $n$ must be even.