Nicely generalizing https://artofproblemsolving.com/community/c6h514287_degrees_gt_3_implies_cycle_not_div_by_3

solved with @Chiaquinha. Consideer the maximal (in vertices) path $P$, with vertices $T_1, T_2, \dots, T_k$. Note that $T_1$ doesn't have connections with any vertices out of the path, due to maximality. Hence, $T_1$ is connected with at least two vertices of $P$, say $T_i, T_j$. Note that we have cycles with sizes $i, j, (j-i +2)$. If exists $p$ with $p \mid i$ , $p \mid j$ , $p \mid j-i+2 \Rightarrow p \mid 2 $, hence the desired $gcd$ only can be $1$ or $2$. To $\gcd(|C_{1}|, |C_{2}|, \dots, |C_{k}|) = 1$, take a regular octogon with each vertex connected to the two adjacents and your antipode. To $\gcd(|C_{1}|, |C_{2}|, \dots, |C_{k}|) = 2$, take a regular hexagon with each vertex connected to the two adjacents and your antipode.

Small note: Here are two other examples of graphs in which the $\rm gcd$ equals $1$ and $2$ respectively: To obtain $\gcd(|C_{1}|, |C_{2}|, \dots, |C_{k}|) = 1,$ take the complete graph on $n$ vertices. Since $n \geq 6$, this graph has a cycle on $3$ and $4$ vertices, hence the $\gcd$ must equal $1$. To obtain $\gcd(|C_{1}|, |C_{2}|, \dots, |C_{k}|) = 2,$ take any bipartite graph on $n$ vertices such that each vertex has degree at least $3$. Since we have a bipartite graph, all its cycles are of even length, hence the $\rm gcd$ can only equal $2$ at this case.

@2above it should be given an example for all $n$. For gcd $1$ take a complete graph while for gcd $2$ take a complete bipartite graph with parts with $n-3$ and $3$ vertices (note that all cycles in any bipartite graph are even).

I wrote an overview on this problem and some others very similar to this one on my blog.

Cute!

answer is $1,2$, hint consider DFS tree