12.101 For which integers $n \geqslant 3$, does there exist a regular $n$-gon in the plane, all of whose vertices are lattice points?
［Solution］(1) Obviously, when $n=4$, there exists a square with integer points.
(2) When $n>6$, if there exists a regular $n$-sided polygon $P$ that meets the requirements, then when each side of it is translated so that one endpoint is the origin, the other endpoint is also an integer point, thus obtaining another regular $n$-sided polygon $P^{\prime}$ with all vertices as integer points. If the side length of $P$ is denoted as $a$, then the side length of $P^{\prime}$ is $a^{\prime}=2 a \sin \frac{\pi}{n}<a$. Repeating this process infinitely, we can obtain an infinite number of strictly decreasing side lengths. However, the distances between integer points not exceeding $a$ are only finitely many, leading to a contradiction.
(3) When $n=3$ or 5, if there exists a regular $n$-sided polygon that meets the requirements, then when the regular $n$-sided polygon $P$ is rotated $90^{\circ}$ around its center three times, a regular 12-sided polygon or a regular 20-sided polygon is obtained, with all vertices as integer points. By (2), this is impossible.
(4) When $n=6$, if there exists a regular hexagon $P$ with integer points, then the 3 non-adjacent vertices of $P$ form a regular triangle with integer points. By (3), this is impossible.
In summary, a regular $n$-sided polygon with all vertices as integer points exists if and only if $n=4$. Constructing a rectangle with the centers of 4 corner grids as vertices, all 28 points lie on the boundary of this rectangle. Drawing 13 lines that do not pass through these points, each line intersects the boundary of the rectangle at most at two points. The 13 lines intersect the boundary of the rectangle at most at 26 points. These intersection points divide the boundary of the rectangle into at most 26 segments. Thus, by the pigeonhole principle, at least one segment contains at least two center points.