The polygon has $n$ vertices. If $n$ is odd or $3$ is not a divisor of $n$ it is easy to see that the frog can jump $n$ times, satisfiying the condition. Suppose $6|n$. Then the frog cannot jump just by jumping clockwise $n$ times or jumping counterclockwise $n$ times. Suppose the frog jumped clockwise $a$ times and counterclockwise $b$ times. Then $a+b=n$. Also, the frog has to end up on the initial vertex, so $n|\left(2a-3b\right)$. The possible values are $2a-3b=-2n,-n,0,n$. In each case, $a=\frac{n}{5}$, $a=\frac{2n}{5}$, $a=\frac{2n}{5}$, $a=\frac{2n}{5}$. So $n$ has to be a multiple of $5$. Therefore $30|n$. Also, if $30|n$, the frog can move by $1\rightarrow 3\rightarrow 5 \rightarrow 2 \rightarrow 4 \rightarrow 6 \rightarrow \cdots \rightarrow \left(n-1\right) \rightarrow 1$. Therefore the set of the possible values is the set of natural numbers which are not a multiple of $6$ or which are a multiple of $30$.

why is 2a-3b = 2n why not anything greater than 2 can frog complete the hops if it goes forward and backward in 2n hops Did not understand the logic.