Claim: $2n = 4 \pmod 8$. I label the vertices with classes $\pmod{2n}$. Suppose the frogs follow the way of jump described (i.e.: after the jump no opposite vertices are both occupied). Sufficiency Consider the following w.o.j.: the frog in vertex $k$ jumps in $k+1$ if $k = 0,1 \pmod 4$, otherwise it jumps in $k-1$. This is easily seen as an example of a good way of jump. [] Necessity Subclaim If, after the jump, a frog lands in $k$, then another frog must land also in $k+n-2$. In fact, the opposite of $k$ is $k+n$ and must be empty. The frog starting in $k+n-1$ must now find itself in $k+n-2$. [] WLOG we can suppose that after the jump, vertex 0 is occupied. Now apply repeatedly the subclaim: every multiple of $n-2$ must be occupied and, this is equivalent to ask that every multiple of $\gcd(2n, n-2)$ must be occupied. Two cases: (even) $2n = 0 \pmod 8$ and (odd) $2n = 2 \pmod 4$. Even case: $\gcd(2n, n-2) = 2$ [because $4 = 2n-2(n-2)$, but $4 \nmid n-2$]. So every even node must be occupied. In particular, $n$ is occupied. But then $0$ and $n$ are opposite vertices both occupied, and this is a contradiction. [] Odd case: $\gcd(2n, n-2) = 1$ [$n-2$ is odd]. So every single node must be occupied and clearly this contradicts the hypotesys. []

We will show that $v_2(n)=1$. Key observations: 1. Any frog has only $2$ vertices to hop onto (since it has $2$ adjacent vertices). So only $2$ vertices are possible for a frog, and also a vertex can have only $2$ possible frogs. 2. There will be $n$ vertices and $2n$ frogs so by 1 each vertex must have exactly $2$ frogs.

and $(y_i, y_{i+1})$ to the even vertices, where $1 \le i \le n$ (and $x_{n+1}=x_1, y_{n+1}=y_1$). 3. By 2, no vertex can be occupied by only $1$ frog. Hence if a vertex $(a,b)$ is chosen, then the 2 vertices of the form $(a,x), (b,y)$ cannot be chosen. Contradiction when $n$ is odd Start with $(x_1,x_2)$ and start moving clockwise. By 3, $(x_2,x_3)$ cannot be chosen. Hence $(x_3, x_4)$ must be chosen. (since we can't leave out $x_3$). Hence $(x_4, x_5)$ cannot be chosen. Hence $(x_5, x_6)$ mst be chosen and so on. So we get the chain (of the vertices we choose): $$(x_1, x_2) \implies (x_3, x_4) \implies (x_5, x_6) \implies \cdots \implies (x_n, x_1)$$which contradicts our point 3 as we cannot chose both $(x_1, x_2)$ and $(x_n, x_1)$. $\blacksquare$ Note: Since $n$ is odd, the antipode of $(x_1, x_2)$ is an even vertex, and so won't bother our chain. Also, since $n$ is odd, hence $(x_n, x_1)$ is a member of our chain.

The figure shows the case when $2n=10$. Contradiction when $v_2(n) \geq 2$ As above, we get $$(x_1, x_2) \implies (x_3, x_4) \implies (x_5, x_6) \implies \cdots \implies (x_{n-1}, x_{n})$$The issue here is that if the antipode of $(x_1, x_2)$ is of the form $(x_{n/2+1}, x_{n/2+2})$, then $n/2+1$ is odd since $n/2$ is even. Hence, the antipode is a part of this chain, a contradiction. $\blacksquare$

The figure shows the case when $2n=8$. Construction when $v_2(n)=1$ It is evident by the above arguments that we must have $2|n$ and $2 \nmid n/2$, i.e. $v_2(n)=1$. We now present our construction for $2n=12$, and in general, the idea is just to choose alternate pairs of vertices. The chosen vertices are marked.

Assume that jumping is over and the condition is satisfied..Label the vertices $A_1 \dots A_{2n}$ with $L$ or $R$ depending on the direction the frog on them has jumped(left or right).Note that at least half of the vertices are empty and every vertex has either 0,1 or 2 frogs,so by PHP,exactly $n$ vertices are empty and so each vertex contains either 0 or 2 frogs.This means that $$ \forall i \leq 2n :A_i \neq A_{i+2} \implies 4|2n \implies \text{n is even}$$that's because if the neighbors of $A_i$ have same labels,then $A_i$ contains exactly one frog,So the labels are $RRLLRR \dots RRLL$ in this order.Note that if every block $RR$ is antipode of a $LL$,the condition would be satisfied,so $n=4k+2$ works and also,if every $RR$ is antipode of a $RR$ then the condition will not satisfied;meaning that $n$ is not divisible by 4,so answer is $n=4k+2$.

Claim: $n\equiv 2 \pmod 4$. There are $2n$ frogs and $2n$ vertices or $n$ pairs of diametrically opposite points. If the frogs end up in more than $n$ vertices, then by PHP, we can find a pair of diametrically opposite vertices each with at least one frog. Number the vertices: $1,2,\dots ,2n$ in anticlockwise direction. We have $2n$ frogs, at most $n$ vertices and at each vertex, at most $2$ frogs (from the neighbours). Thus $n$ vertices have exactly $2$ frogs. Without loss of generality, vertex $1$ has $2$ frogs and frog from vertex $1$ jumped to vertex $2$. Notice that each vertex contains either two frogs or zero. Now vertex $2$ also has $2$ frogs or frog from vertex $3$ jumped to vertex $2$. As frog from vertex $2$ jumped to vertex $1$, frog from vertex $4$ jumped to vertex $5$ and so on. Also notice that frog from vertex $1$ did not jump to vertex $2n$ thus vertex $2n$ has $0$ frogs. Concluding we get that vertex $k$ has two frogs if and only if $k\equiv 1,2\pmod 4$. We have that vertex $2n$ has $0$ frogs or $2n\equiv -1,0\pmod 4$. Thus $n$ is even, say $n=2m$. Mark 'X' on a vertex if it has 2 frogs and 'O' if it has 0. If $n+1$ labelled vertex is marked $X$ then we get two diametrically opposite points with at least one frog on them. Thus we need $n+1\equiv 3,0\pmod 4$ or $2m\equiv 2,3\pmod 4$ or $m$ is odd. Thus $n = 4m+2$. Construction for $n \equiv 2\pmod 4$: The frogs on vertex $4k+1$ and $4k+2$ swap after jump. Frog on $4k+3$ jump on the left vertex or $4k+2$ and frog on vertex $4k+4$ jump on the right vertex or $4k+5$.

$n=4k+2$ First note that after the move each vertex has atmost 2 frogs since there are two neighbouring vertices.Call a vertex rich if it has frogs after the move and poor otherwise.Since no two opposite vertices are rich there are atmost $n$ vertices but since each vertex can accommodate atmost $2$,we conclude all rich vertices have $2$ frogs each Claim $1$:$n$ is even. Suppose not. Consider a bipartite graph of rich and poor vertices.Draw an edge between $2$ vertices if they are adjacent.So every vertex has degree $2$.Notice that every poor vertex has to be connected to a rich vertex,since every frog is jumping from its place.Also a poor vertex cannot be connected to $2$ rich vertices which will imply that it is jumping at $2$ distinct places.Therefore every poor vertex is connected to exactly one poor vertex and every rich vertex is connected to exactly one rich vertex.But this means there are $\frac {n}{2}$ edges among poor vertices,contradiction to our assumption. Claim $2$:$n\neq 4k$ Now a poor vertex $a_1$ is connected to a rich vertex $b_1$which is connected to rich vertex $b_2$ which is connected to poor vertex $a_2$.So they must lie in this order on the polygon(say in clockwise direction).Now to the right of $a_2$ there is a poor vertex $a_3$.So this process of formation of quadruple of $(a_{2k+1},a_{2k+2},b_{2k+1},b_{2k+2})$ will continue.But since the vertex numbers of opposite vertices are congruent mod $4$,we get that both are rich or both are poor.This is a contradiction. For $n=4k+2$ the construction is simple.$4m+1$ and $4m+3$ jump to $4m+2$ and $4m+2$ and $4m+4$ jump on $4m+3$ where $m$ takes values from $0$ to $2k$

Note : I’ll be sometimes calling “inhabited” vertices R or red, similarly with “uninhabited” (blue) We claim that $n=4k+2$ is the only possible value. Call a vertex “inhabited” if there’s at least one frog on it (after the algorithm is applied), and call a vertex “uninhabited” if not. Claim 1: every vertex has 2 or 0 frogs on it Proof : initially, there are $2n$ frogs. Now, $a_{1}+a_{2}+ \dots + a_{i} = 2n \leq 2i \leq 2n$. This occurs when $i=n, $ and there are two frogs on all inhabited vertices $\Box$ Claim 2: at least 2 inhabited vertices are adjacent. Proof : The frog on any inhabited vertex must jump to an adjacent vertex. claim 3: the vertices must occur as $RRBBRRBB \dots$ Proof : Let the vertices be labelled in the format $A_i$, such that the $i$s increase by one as we move clockwise. (Diagram below). Let $A_1, A_2$ be adjacent R’s. $A_3$ should be $B$ as $A_2$s frog must have gone to $A_1$ and $A_3$s frog to $A_2$. $A_4$s frog going to $A_3$ isnt possible coz then it has only 1 frog. Similarly, $A_4$ is uninhabited too. Now $A_4$ s frog must have gone to $A_5$ so it is inhabited. Similarly $A_5$s frog went to $A_6$ so it is inhabited too. $A_{4n}$ and $A_{4n-1}$ have to be uninhabited coz $A_1$s frog went to $A_2$. So basically, the string $RRBB$ is repeating. this implies the thing has $4k$ vertices (as if there are $4k+2$, it will end with $AAAA$, which as we said is not possible). Now, let $\text{number of vertices}=4t$. Note that $A_{2t+1}, A_{2t+2}$ are B. We see that $A_i$ is R if $i$ is $1,2 \mod 4$, B otherwise. So $2t+1 \equiv{3} \pmod{4} \implies t \equiv{1} \pmod{2}$. So number of vertices $=2n=4t=4(2k+1) \implies n \equiv{2} \pmod{4}$ Construction: label vertices as $a_i$s mod 4, that is $1,2,3,4,1,2,3,4, \dots$. Consider consecutive vertices $4,1,2,3$. Transfer frogs from $4,2$ to $1$, and from $1,3$ to $2$ for all $\blacksquare$

Solved with None since I am lonely and just got kicked out of my last Math group. We claim is that the answer is just $n \equiv 2\pmod 4$. Actually,one of our insights is that the simultaneous condition is not of any concern.We can just move the frogs one by one. So,we just place a $1$ at each of the vertices and then $1'$ for a frog that has moved. Now, we just start moving with the first frog and see that the diametrically opposite vertex just gets fixed and we circle it. Now, the $1s$ adjacent to it have to move in a specific direction and thus we gain some insight. With some casework by observing the patterns in the occurrence of $1's$ ,circles and empty points ,we are done.

The answer is $n \equiv 2 \pmod 4$. All the congruences below are taken modulo $2n$. After the frogs move, we color each vertex which has at least one frog. CLAIM: there are exactly $n$ colored vertices, each of them has exactly two frogs. Proof: First notice that the total number of diameters which connect the vertices of the 2n-gon is $n$, and each diameter has at most one colored vertex, hence $n$ is the maximum number of colored vertices. Also it is easy to see that each colored vertex has at most two frogs, hence the claim is proved. Let $A_1A_2...A_{2n}$ be the 2n-gon, and take the indices modulo $2n$. Assume that $A_i$ is colored, then $A_i$ has two frogs from $A_{i-1}$ and $A_{i+1}$. Now it is easy to see that $A_{i-1}$ or $A_{i+1}$ is colored. Assume WLOG that $A_{i+1}$ is colored, then we can see a pattern of colored vertices, that is, each one of $A_{i+4k}$ and $A_{i+1+4k}$ is colored. $\boxed A$ Since $A_{i-1}$ isn't colored, then: $$i-1 \not\equiv{i+1+4k} \Rightarrow 2\not\equiv{4k} \Rightarrow 2\mid n$$$\boxed B$ $A_{i+n}$ is the antipode of $A_i$, then: $$i+n \not\equiv i+4k \Rightarrow n\not\equiv 4k \Rightarrow 4 \nmid n$$Hence $v_2(n)=1$, and this actually works due to $A$ and $B$. $\blacksquare$

After the jumps, mark the points occupied by frogs with O, and the empty points with X. Clearly for each diagonal of the polygon that passes through its circumcenter, exactly one of its endpoints is an O and the other is an X. We want to mark the vertices of this polygon in this manner, and also such that each vertice is adjacent to an O. Now, define a "block" to be the maximum-lengthed continuous string of the same letter present in the polygon. Clearly for each block of Xs, its length must be at most $2$ as otherwise one of the vertices in the middle of the block is not adjacent to an O. Therefore each block of Xs have either length $1$ or $2.$ But if it has length $1,$ it is surrounded by Os on two sides, and therefore the opposite side of the polygon has an O surrounded by two Xs. But then this O vertice is not adjacent to any other O vertices. Therefore the length of an X block can only be $2.$ Now, for O blocks, their lengths must clearly be at least $2.$ But if they have length greater than $2,$ the opposite side of the polygon must have a block of Xs with length at least $3,$ a contradiction. Thus O blocks have length $2.$ Hence, denote the vertices as $P_1, P_2, \cdots, P_{2n}.$ WLOG $P_1, P_2$ are O. Then $P_3, P_4$ are X, etc. until we cycle back. Then $P_{4k+1}, P_{4k+2}$ are Os while $P_{4k+3}, P_{4k+4}$ are Xs and thus $2n \equiv 0 \pmod 4.$ However, recall that opposites cannot equal each other, hence $P_{1} \neq P_{n+1}$ so $\boxed{n \equiv 2 \pmod 4}.$ Clearly this works.