1. Let $d_{1}, d_{2}, \cdots, d_{k}$ be all the divisors of the positive integer $n$, $1=d_{1}<d_{2}<d_{3}<\cdots<d_{k}=n$. Find all $n$ such that $k \geqslant 4$ and $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}=n$.
Number, but the sum of the squares of four odd numbers is even, which cannot equal $n$. Contradiction.
Thus, $n$ is even, $d_{2}=2$.
If $n$ is a multiple of 4, then $4 \in\left\{d_{3}, d_{4}\right\}$. Among the squares of the 4 factors, there are already two $\left(2^{2}\right.$ and $\left.4^{2}\right)$ that are multiples of 4, and one is 1. When another factor is odd, its square is congruent to 1 modulo 4; when even, its square is divisible by 4. Therefore, the sum of the squares of these 4 factors is not a multiple of 4, leading to a contradiction again.

Thus, $n=2 m, m$ is odd. $d_{3}$ is the smallest prime factor of $m$. Since $1^{2}+2^{2}+d_{3}^{2}+d_{4}^{2}=n$ is even, $d_{4}$ must be even, so $d_{4}=2 d_{3}$.
$$
n=1+2^{2}+d_{3}^{2}+4 d_{3}^{2} \Rightarrow 5 \text { is a divisor of } n.
$$

Excluding $d ; 4$, so $d \geqslant 6$, thus $d_{3}=5, d_{4}=10, n=1+2^{2}+5^{2}+10^{2}=130$.