Five. (20 points) Find all positive integers $n$ such that $n=$ $p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}$, where $p_{1}, p_{2}, p_{3}, p_{4}$ are the four smallest distinct positive divisors of $n$.
If $n$ is odd, then all factors of $n$ are odd, i.e., $n \not\equiv 0(\bmod 4)$.
But $n=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2} \equiv 0(\bmod 4)$, which is a contradiction.
Therefore, $21 \nmid n$.
If $4 \mid n$, then $p_{1}=1, p_{2}=2$.
Thus, $n=1+0+p_{3}^{2}+p_{4}^{2} \not\equiv 0(\bmod 4)$, which is a contradiction.
Therefore, $4 \nmid n$.
Hence, $\left\{p_{1}, p_{2}, p_{3}, p_{4}\right\}$
$=\left\{1,2, p_{3}, p_{4}\right\}$ or $\left\{1,2, p_{3}, 2 p_{3}\right\}$.
If $\left\{p_{1}, p_{2}, p_{3}, p_{4}\right\}=\left\{1,2, p_{3}, p_{4}\right\}$, this is impossible.
Thus, $\left\{p_{1}, p_{2}, p_{3}, p_{4}\right\}=\left\{1,2, p_{3}, 2 p_{3}\right\}$, i.e., $n=5\left(1+p_{3}^{2}\right)$.
So, $p_{3}=5$.
Therefore, $n=130$.