49th Mathematical Olympiad in Ukraine

Let us denote the number of boys at the party by $x$ and the number of girls by $y$. Then from the problem statement we can get the following equality: $2xy=4(3y+2x)$, which is equivalent to the equation $(x-6)(y-4)=24$. Now it is enough to consider all divisors of $24$ and to get all possible $x$ and $y$, among which the greatest value of $(x+y)$ is achieved in the two cases: $$\{\begin{array}{l}x-6=1\\ y-4=24\end{array}\text{ or }\{\begin{array}{l}x-6=24\\ y-4=1\end{array}$$ or this greatest value is: $$x+y=7+28=30+5=35.$$