An easy one there are 36 solutions. 9,10,11,12,13,14,15,16, 18,19,20,21,22,23,24, 27,28,29,30,31,32, 36,37,38,39,40, 45,46,47,48, 54,55,56, 63,64, 72.

Sorry to bring this post back up but the current solution seems very vague (in fact, it's not quite a solution at all). Also, I'm interested in the problem and I want to bring this post back to life so that it can have another chance of someone finding and posting a solution.

It probably hasn't been solved or gotten a detailed solution because it's rather ugly casework without any neat insights to be had. Suppose there are $k$ groups of males and $17 - k$ groups of females; WLOG $k \geq 9$. Also suppose that all the groups have $m$ or $m+1$ members. Then $mk \leq (m+1)(17-k)$, or $(2m+1)(2k-17) \leq 17$. Since $m \geq 1$, we have $2k - 17 < 6$, so $9 \leq k \leq 11$. $k = 11$? Then only $m = 1$ works. We have at least 11 males and at most 12 females. So $n = 11, 12$ work. $k = 10$? Then only $m = 1$ or $m = 2$ works. For $m = 1$, we have at least 10 males and at most 14 females. So $n = 10$ through $14$ work. For $m = 2$, we have at least 20 males and at most 21 females. So $n = 20, 21$ work. $k = 9$? Then $m$ can be as high as 8. For a given $m$, we have at least $9m$ males and at most $8(m+1)$ females. This gives the range of solutions seen in post #2 which I won't recopy here. It turns out that this set of solution subsumes all the previous cases, so that is the complete list of answers.