1978 AMC 12 Problem 30

Let $k$ be the number of mixed matches won by women and count all wins by women

Use $0\le k\le2n^2$ together with the required $\frac{7}{12}$ share of all match wins

There are $\frac{3n(3n-1)}{2}$ matches, so women must win $\frac{7n(3n-1)}{8}$ matches. If women win $k$ of the $2n^2$ mixed matches, then $$\frac{n(n-1)}2+k=\frac{7n(3n-1)}8,$$ giving $k=\frac{n(17n-3)}{8}.$ The bound $k\le2n^2$ forces $n\le3.$ For $n=1,$ $2,$ this formula is not an integer, while $n=3$ gives $k=18,$ which is possible. Thus $n=3,$ which is not among the listed numerical choices.

Therefore, the correct answer is E.