Colors are irrelevant. Let us consider the groups of contiguous green children, having gaps of $\delta\geq 1$ between them. Then all the children sitting in the $\alpha$ gaps with $\delta = 1$ will raise their hand, while among the children sitting in the $\beta$ gaps with $\delta \geq 2$, exactly two will raise their hand. Thus $\alpha + 2\beta = 20 + 25 = 45$, an odd number, hence $\alpha$ must be odd, and so of value at least $1$.

Shu wrote: Red, blue, and green children are arranged in a circle. When a teacher asked the red children that have a green neighbor to raise their hands, $20$ children raised their hands. When she asked the blue children that have a green neighbor to raise their hands, $25$ children raised their hands. Prove that some child that raised her hand had two green neighbors. Note. This contest concluded on July 19. Results of the contest have already been published. So it's okay to discuss this problem. Because $25-20$ is odd and if there weren't people with two green neighbours, the difference would be odd, look to the way between two neighbouring green people (no green between them): $GBG$ and $GRG$ doesn't happen $GB...BG$ means $[B]-[R]$ has increased with $2$ and similar for $GR...RG$ with $-2$ $GR..BG$ and $GB..RG$ gives the difference didn't change. Hence it would imply the diffr. is even. For which grade is that? ps: it seems mavro.. has again a better mathwritten sol.

We can ignore the sequence when there are at least 2 neighboring green children such that we consider this sequence one green children. Let's consider this sequence $G....G.....G....G....$ (we imagine that this is a circle). Assume that all the children that raised his hand had one green neighbors. $r$-the number of red children who has a green neighbor. So $r$ is the number of red children who raised their hand. $b$-the number of blue children who has green neighbor. So $b$ is the number of blue children who raised their hand. We have $r+b=55$ which is odd, but $r+b$ $is$ $the$ $number$ $of$ $green$ $children$ *$ 2 $ . So we have $55=even$ contradiction.