A certain frog that was placed on a vertex of a convex polygon chose to jump to another vertex, either clockwise skipping one vertex, either counterclockwise skipping two vertexes, and repeated the procedure.
If the number of jumps that the frog made is equal to the number of sides of the polygon, the frog has passed through all its vertexes and ended up on the initial vertex, what´s the set formed by all the possible values that this number can take?


[i]Andrei Eckstein[/i]
1. Let the polygon have \( n \) vertices. The frog can jump either clockwise skipping one vertex or counterclockwise skipping two vertices. The frog makes \( n \) jumps and ends up on the initial vertex after passing through all vertices.

2. We need to determine the set of possible values for \( n \) such that the frog's jumps satisfy the given conditions.

3. If \( n \) is odd or \( 3 \) is not a divisor of \( n \), the frog can jump \( n \) times and satisfy the condition. This is because the frog can visit all vertices and return to the starting vertex without any restrictions.

4. Suppose \( 6 \mid n \). Then the frog cannot jump just by jumping clockwise \( n \) times or jumping counterclockwise \( n \) times. We need to consider a combination of both types of jumps.

5. Let \( a \) be the number of clockwise jumps and \( b \) be the number of counterclockwise jumps. Then we have:
   \[
   a + b = n
   \]
   Additionally, the frog must end up on the initial vertex, so:
   \[
   n \mid (2a - 3b)
   \]

6. The possible values for \( 2a - 3b \) are \( -2n, -n, 0, n, 2n \). We solve for \( a \) in each case:
   - For \( 2a - 3b = -2n \):
     \[
     2a - 3(n - a) = -2n \implies 2a - 3n + 3a = -2n \implies 5a = n \implies a = \frac{n}{5}
     \]
   - For \( 2a - 3b = -n \):
     \[
     2a - 3(n - a) = -n \implies 2a - 3n + 3a = -n \implies 5a = 2n \implies a = \frac{2n}{5}
     \]
   - For \( 2a - 3b = 0 \):
     \[
     2a - 3(n - a) = 0 \implies 2a - 3n + 3a = 0 \implies 5a = 3n \implies a = \frac{3n}{5}
     \]
   - For \( 2a - 3b = n \):
     \[
     2a - 3(n - a) = n \implies 2a - 3n + 3a = n \implies 5a = 4n \implies a = \frac{4n}{5}
     \]
   - For \( 2a - 3b = 2n \):
     \[
     2a - 3(n - a) = 2n \implies 2a - 3n + 3a = 2n \implies 5a = 5n \implies a = n
     \]

7. From the above, we see that \( n \) must be a multiple of \( 5 \). However, for the frog to visit all vertices and return to the starting vertex, \( n \) must also be a multiple of \( 6 \). Therefore, \( n \) must be a multiple of \( 30 \).

8. If \( 30 \mid n \), the frog can move in a sequence such as \( 1 \rightarrow 3 \rightarrow 5 \rightarrow 2 \rightarrow 4 \rightarrow 6 \rightarrow \cdots \rightarrow (n-1) \rightarrow 1 \).

9. Therefore, the set of possible values for \( n \) is the set of natural numbers that are not multiples of \( 6 \) or are multiples of \( 30 \).

The final answer is \( \boxed{ \{ n \in \mathbb{N} \mid 6 \nmid n \text{ or } 30 \mid n \} } \)