$\begin{tabular}{||c c c||} \hline Time & girls & boys\\ [0.5ex] \hline\hline \text{Initially} & m & n\\ \hline 1st stop & 2m/3 & n+m/3\\ \hline \text{2nd stop} & 2m/3 +(1/3)(n+m/3) & (2/3)(n+m/3)\\ \hline \end{tabular}$ $2m/3 +(1/3)(n+m/3) =(2/3)(n+m/3)+2$ $5m-3n=18$ $(2/3)(n+m/3) =m$ $m=(6/7)n$ $m=12, n=14$ 12 girls and 14 boys

Let $b$ be the original number of boys and $g $ be the original number of girls. After the first stop, we have $b+\frac{g}{3}$ boys and $\frac{2g}{3}$ girls. After the second stop, there are $\frac23\left(b+\frac g3\right)$ and $\frac{2g}{3}+\frac13\left(b+\frac g3\right)$ girls. We have that $\frac23\left(b+\frac g3\right)+2=\frac{2g}{3}+\frac13\left(b+\frac g3\right)$ and $\frac23\left(b+\frac g3\right)=g.$ Solving these, we obtain $b=14$ and $g=12.$