Answer: $65$. Example: Take $35$ boys and $30$ girls, which are $B_1,...,B_{35},G_1,...,G_{30}$. We divide $35$ boys into $5$ groups where all group has $7$ members and $30$ boys into $5$ groups where all group has $6$ members. Call these groups $A_1,...,A_5$ and $B_1,...,B_5$ respectively. Connect $A_i$ with $B_i$ and $B_{i+1}$. Then all members in $A_i$ has degree $12$ and all members in $B_i$ has degree $14$. Proof: Let there be $b$ boys and $g$ girls. Construct a table with $b$ rows and $g$ columns. If $b_i$ beats $g_j$, then we colour $i.$ row $j.$ column into blue and red otherwise. There are $bg$ squares on the table and $\geq 12b$ blue squares, $\geq 21g$ red squares. Thus $bg\geq 12b+21g\iff g(b-21)\geq 12b$ We have $b>21$ so if we assume that $b+g\leq 64\iff g\leq 64-b,$ then we would have $12b\leq g(b-21)\leq (64-b)(b-21)=-b^2+85b-64.21\iff b(73-b)\geq 64.21$ Also $b(73-b)\leq 37.36\implies 37.36\geq b(73-b)\geq 64.21\implies 111=37.3\geq 16.7=112$ which is impossible.