4. On each vertex of a regular $2 n$-sided polygon inscribed in a circle $(n \geqslant 2)$, there is a frog. At a certain moment, they all jump simultaneously to one of the adjacent vertices (it is allowed for multiple frogs to land on the same vertex), which is called a jump method. Try to find all possible values of $n$ such that for this $2 n$-sided polygon, there exists a jump method where after jumping, the line connecting any two different vertices with frogs does not pass through the center of the circle.
4. The answer is $n \equiv 2(\bmod 4)$.

First, construct a jumping method for $n \equiv 2(\bmod 4)$:
Let $n=4 k+2$, then $2 n=8 k+4$. Initially, choose any frog and number the vertex it is on as 1, then number the remaining $2 n-1$ frogs' vertices in a clockwise direction as $2,3, \cdots, 8 k+4$. For the frog at vertex $j$, if $j \equiv 1, 2(\bmod 4)$, the frog jumps clockwise; if $j \equiv 3, 4(\bmod 4)$, the frog jumps counterclockwise.

According to the above rules, after jumping, only the vertices numbered $2,6,10, \cdots, 8 k-2,8 k+2 ; 3,7,11, \cdots, 8 k-1,8 k+3$ have frogs, while the vertices numbered $1,5,9, \cdots, 8 k-3,8 k+1 ; 4,8,12, \cdots, 8 k, 8 k+4$ have no frogs. At this point, the difference in the numbers of any two vertices with frogs is not $n=4 k+2$, so the line connecting any two vertices with frogs does not pass through the center of the circle.
On the other hand, prove that $n \neq 2(\bmod 4)$ is impossible:
Suppose the $2 n$ frogs' vertices are numbered $1,2, \cdots, n, \cdots, 2 n$ in a clockwise direction. If all frogs jump in the same direction (clockwise or counterclockwise), this is a global rotation, and there must be two frogs whose vertices' line passes through the center of the circle. Therefore, there must be a frog that jumps in the opposite direction. Suppose the frog at vertex $l$ jumps counterclockwise, then the frog at vertex $l+n-2$ cannot jump clockwise, it can only jump counterclockwise (otherwise, the vertices numbered $l-1$ and $l+n-1$ will both have frogs, and the line connecting these two vertices will pass through the center of the circle). By induction, the frogs at vertices $l+m(n-2)$ jump counterclockwise $(m=1,2, \cdots)$.
(1) If $n$ is odd, then $(n-2,2 n)=1$, so $l+m(n-2)(m=1,2, \cdots)$ covers the complete residue system of $2 n$, leading to all frogs jumping counterclockwise, which means there must be two frogs whose vertices' line passes through the center of the circle;
(2) If $n \equiv 0(\bmod 4)$, let $n=4 k$, then $(n-2,2 n)=2$, indicating that $l+m(n-2)$ covers all residues of $2 n$ with the same parity as $l$, i.e., $l, l+2, l+4, \cdots, l+8 k$. In this case, both $l$ and $l+4 k$ jump counterclockwise, so after jumping, there must be two frogs whose vertices' numbers are $l-1$ and $l+4 k-1$, and the line connecting these two vertices passes through the center of the circle.
In summary, the required $n$ is $n \equiv 2(\bmod 4)$.