Proving that there is at least one person, both of whose neighbors are girls. $25$ boys and $25$ girls are sitting around a table.
Prop: There is at least one person, both of whose neighbors are girls.
Pf: Assume that $B$ sits on the first chair. Further assume that $n_i\geq 0$ denotes the number of $B$'s we're adding at $i$th step. Choose $24>n_1\geq 0$ such that $24-n_1>1$. Then, there has to be some $G$ such that $B....BG(B/G)B$. If there does not exist $n_2\geq 0$ such that $23-(n_1+n_2)>1$, then $B......BG(B/G)G$ trivializes the assertion. If there does not exist $n_3\geq 0$ such that $22-(n_1+n_2+n_3)>1$, then $B....BG(B/G)B....BG(B/G)G$ trivializes the assertion. Then, as $n\rightarrow 23$, $1-(\sum_{1\leq i\leq n}n_i)>0$, and so, $\sum_{1\leq i\leq n}n_i\rightarrow 0$. In which case we'll end up with $BG(B/G)BGBG(B/G)BGBG(B/G)BG...$
Here, $B/G$ means $B$ or $G$. Is my proof correct?
 A: As Mark Bennet pointed out, your solution is problematic, because it does not rely on the fact that there is an odd number of children of the same gender.
Something between an extended hint and a roadmap in how to take advantage of the odd parity.

*

*Color all the chairs alternately black and white. Why do the girls have a majority on one of the colors?

*Assume that there are more girls than boys sitting on a black chair. So at least 13 girls sit on a black chair. If we leave the white chairs out of the reckoning, why does it follow that there is a girl on two adjacent black chairs?

*Why does the conclusion of the previous bullet solve your question?

A: Say we had $n>1$ of each and suppose we had an arrangement in which no child was surrounded by girls.
Choose some girl with a male counterclockwise neighbor (easy to see that such must exist).  Then, starting from her, we can describe the cyclic pattern as $$\{G^{a_1}B^{b_1}\cdots G^{a_k}B^{b_k}\}$$
Now, each $a_i$ is either $1$ or $2$ (as we can't have a block of three or more girls). And each $b_i$ is $>1$.  Since $\sum a_i=\sum b_i=n$ we see that each of $a_i, b_i$ must be $2$.  But that forces $n$ to be even.  In your case $n$ is odd so we are done.
A: First you put down the 25 boys and name them by $x_1,...,x_{25}$ in the clockwise order, say.
Suppose each boy is next to at most one girl.
Without loss of generality, we may assume a girl sits on the right of $x_1$ (otherwise we can reverse the order and rename boys.) Then you go clockwise and count how many girls you meet. If the girl next to $x_{i+1}$ sits on the left, then you already met her when you consider $x_i$. Therefore to arrange all the 25 girls, you must have each of them sit right to a $x_i$. But then on the left of $x_1$, there is a girl (sitting right to $x_{25}$). Contradiction.
