Let $g_1,g_2, \ldots, g_{2x}$ be th girls and $b_1,b_2, \ldots, b_x$ be the boys. Let $G,B$ be the total wins of girls and boys, respectively. In total, $\binom{3x}{2}$ matches were played, so we have in total $\binom{3x}{2}$ wins. So, $G+B=\dfrac{(3x-1)3x}{2}$. Also, we know that $\dfrac{G}{B}=\dfrac{7}{9}$, so $\dfrac{16G}{7}=\dfrac{(3x-1)3x}{2}$. The games played by two girls are $2x^2-x$, so $G \geqslant 2x^2-x \Rightarrow \dfrac{(3x-1)3x}{2} \geqslant \dfrac{32x^2-16x}{7} \Rightarrow x \leqslant 11$. But, we must have $G=\dfrac{21x(3x-1)}{32}$, so $32 \mid 3x^2-x$ and since $x \leqslant 11$, we can easily obtain $x=11$. We can now construct an example to show that this value is attainable, so $\boxed{33}$ players took part.