$n>2$, let $d$ be the largest integer such that $d|n!$ and $d^2<n!$,is it true that $d$ is even?
I checked this for $3\leq n\leq 35$,the highest power of $2$ that divides $d$ is $1, 2, 1, 3, 1, 6, 6, 1, 2, 5, 10, 3, 5, 7, 6, 8, 7, 7, 5, 8, 2, 13, 13, 6, 4, 8, 2, 5, 2, 4, 9, 13, 21.$
Let $$a(p)=\sum_{i=1}^{\infty}\lfloor \frac{n}{p^i}\rfloor$$ then $$n!=\prod_{p\leq n}p^{a(p)}.$$ If $$d=\prod_{p\leq n}p^{b(p)}<\sqrt{n!}$$ then $$\log d=\sum_{p\leq n}{b(p)\log p}<\log \sqrt{n!}.$$
This is like how to choose a pile of stones whose total weight is closest to the capacity of a bag, will the smallest stone be chosen?
In terms of probability, the probability of $b(p)=0$ is $\frac{1}{1+a(p)}$, then the probability of $b(2)>0$ for all $n>2$ is about $$\prod_{n=3}^{\infty}\frac{a(2)}{1+a(2)}\approx \prod_{n=3}^{\infty}\frac{n}{1+n} =0,$$ so mayby there are any counterexamples?