Let $n$ be a positive integer such that $n \geq 2$. We aim to find all possible values of $n$ for which there exists a way of jump in a regular $2n$-gon such that the line connecting any two distinct vertices having frogs on it after the jump does not pass through the circumcenter of the $2n$-gon.

### Key Observations:
1. Each frog has only 2 adjacent vertices to jump to, and each vertex can have at most 2 frogs.
2. There are $2n$ frogs and $n$ vertices, so each vertex must have exactly 2 frogs after the jump.

### Contradiction When $n$ is Odd:
- Label the vertices $v_1, v_2, \ldots, v_{2n}$.
- Assign pairs $(x_i, x_{i+1})$ to odd vertices and $(y_i, y_{i+1})$ to even vertices.
- Starting with $(x_1, x_2)$ and moving clockwise, we must choose $(x_3, x_4)$, then $(x_5, x_6)$, and so on, forming a chain:
$$(x_1, x_2) \implies (x_3, x_4) \implies (x_5, x_6) \implies \cdots \implies (x_n, x_1).$$
- This chain contradicts the requirement that no vertex can be occupied by only one frog, as both $(x_1, x_2)$ and $(x_n, x_1)$ cannot be chosen simultaneously.

### Contradiction When $v_2(n) \geq 2$:
- For $n$ such that $v_2(n) \geq 2$, the chain formed includes the antipode of $(x_1, x_2)$, leading to a contradiction.

### Construction When $v_2(n) = 1$:
- For $n$ such that $v_2(n) = 1$, we can construct a valid jump by choosing alternate pairs of vertices.
- This ensures that no line connecting two distinct vertices with frogs passes through the circumcenter.

Hence, the only possible values of $n$ are those for which $v_2(n) = 1$. This means $n$ must be of the form $2^k \cdot m$ where $k = 1$ and $m$ is an odd integer.

The answer is: \boxed{2^k \cdot m \text{ where } k = 1 \text{ and } m \text{ is an odd integer}}.