Three, (16 points) Let $n$ be a positive integer, and $d_{1}<d_{2}<$ $d_{3}<d_{4}$ be the 4 smallest consecutive positive integer divisors of $n$. If $n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}$, find the value of $n$.
If $n$ is odd, then $d_{1} 、 d_{2} 、 d_{3} 、 d_{4}$ are all odd. Hence $n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2} \equiv 1+1+1+1 \equiv 0(\bmod 4)$. Contradiction.
If $4 \mid n$, then $d_{1}=1, d_{2}=2$. By $d_{i}^{2} \equiv 0$ or $1(\bmod 4)$, we have $n \equiv 1+0+d_{3}^{2}+d_{4}^{2} \neq 0(\bmod 4)$. Also a contradiction.
Therefore, $n=2\left(2 n_{1}-1\right), n_{1}$ is some positive integer, and the tuple $\left(d_{1}, d_{2}, d_{3}, d_{4}\right)=(1,2, p, q)$ or $(1,2, p, 2 p)$, where $p, q$ are odd primes.
In the first case, $n=1^{2}+2^{2}+p^{2}+q^{2} \equiv 3(\bmod 4)$. Contradiction.
Thus, it can only be $n=1^{2}+2^{2}+p^{2}+(2 p)^{2}=5\left(1+p^{2}\right)$. Hence $51 n$.
If $d_{3}=3$, then $d_{4}=5$, which will return to the first case, so it can only be $d_{3}=p=5$, then $n=1^{2}+2^{2}+5^{2}+10^{2}=130$.

It is easy to verify that the 4 smallest consecutive positive divisors of 130 are $1,2,5,10$, satisfying the condition.
Therefore, $n=130$.