308. A line is drawn through the center of a regular $n$-sided polygon inscribed in a unit circle. Find the sum of the squares of the distances from the vertices of the $n$-sided polygon to this line.
308. Note that if we consider a system of vectors originating from the center of a regular $n$-gon and ending at its vertices, the sum of these vectors is zero. Indeed, if we rotate all these vectors by an angle of $2 \pi / n$, their sum does not change, but on the other hand, the vector equal to their sum will rotate by the same angle. Therefore, the sum of the projections of these vectors onto any axis is also zero.

Let's return to the problem. If $\varphi$ is the angle between the given line (denote it as $l$) and one of the vectors, then the other vectors form angles $\varphi+\frac{2 \pi}{n}, \varphi+2 \frac{2 \pi}{n}, \ldots, \varphi+(n-1) \frac{2 \pi}{n}$. The square of the distance from the $k$-th vertex to $l$ is $\sin ^{2}\left(\varphi+k \frac{2 \pi}{n}\right)^{2}=\frac{1}{2}\left(1-\cos \left(2 \varphi+k \frac{4 \pi}{n}\right)\right)$. However, the values $\cos \left(2 \varphi+k \frac{4 \pi}{n}\right)$ can be considered as the projections of a system of $n$ vectors onto $l$, which form angles $2 \varphi+k \frac{4 \pi}{n}(k=0,1, \ldots, n-1)$ with $l$. When $n$ is odd, these vectors form a regular $n$-gon, and when $n$ is even, it will be a twice repeated $\frac{n}{2}$-gon. Answer: $\frac{n}{2}$.