Three. (50 points) Let $\tau(x)$ denote the number of positive divisors of the positive integer $x$. Find all positive integers $n$ such that
$$
\tau(n!) \nmid n!\text {. }
$$
When $n \leqslant 7$, it is known through calculation that only 3 and 5 satisfy the condition.
Below, we prove that when $n \geqslant 8$, $\tau(n!) \mid n!$.
Let $2=p_{1}<p_{2}<\cdots<p_{l} \leqslant n$ be all the primes not greater than $n$,
$$
v_{p_{i}}(n!)=\sum_{\alpha=1}^{\infty}\left[\frac{n}{p_{i}^{\alpha}}\right]<\sum_{\alpha=1}^{\infty} \frac{n}{p_{i}^{\alpha}}=\frac{n}{p_{i}-1} .
$$

Then $1+v_{p_{i}}(n!) \leqslant\left\lceil\frac{n}{p_{i}-1}\right\rceil(1 \leqslant i \leqslant l)$.
For any prime $p \leqslant n$, prove
$$
v_{p}(\tau(n!)) \leqslant v_{p}(n!) \text {. }
$$

In fact,
$$
\begin{array}{l}
v_{p}(\tau(n!))=\sum_{i=1}^{l} v_{p}\left(1+v_{p_{i}}(n!)\right) \\
\leqslant \sum_{i=1}^{l}\left[\log _{p}\left[\frac{n}{p_{i}-1}\right]\right] \\
=\sum_{k=1}^{\infty}\left[\sum_{\left.\log _{p}\left[\frac{n}{p_{i}-1}\right]\right] \geqslant k} 1,\right. \\
V_{p}(n!)=\sum_{\alpha=1}^{\infty}\left[\frac{n}{p^{\alpha}}\right] .
\end{array}
$$

We now prove: $\sum_{\left[\log _{p}\left[\frac{n}{p_{i}-1}\right]\right] \geq k} 1 \leqslant\left[\frac{n}{p^{k}}\right]$.
Notice,
$$
\begin{array}{l}
{\left[\log _{p}\left[\frac{n}{p_{i}-1}\right]\right] \geqslant k} \\
\Leftrightarrow \log _{p}\left[\frac{n}{p_{i}-1}\right] \geqslant k \\
\Leftrightarrow\left\lceil\frac{n}{p_{i}-1}\right] \geqslant p^{k} \\
\Leftrightarrow p_{i} \leqslant\left\lceil\frac{n}{p^{k}-1}\right\rceil .
\end{array}
$$

Thus, it suffices to prove that for any $t \geqslant 1$, the number of primes not exceeding $\left\lceil\frac{n}{t}\right\rceil$ is no greater than $\left\lceil\frac{n}{t+1}\right]$. This conclusion is easy to prove.
Therefore, for any prime $p \leqslant n$, we have $v_{p}(\tau(n!)) \leqslant v_{p}(n!)$.
Thus, when $n \geqslant 8$, $\tau(n!) \mid n!$.

Hence, all $n$ that satisfy the condition are 3 and 5.