4.32 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 
\[
d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}=n.
\]
(6th Balkan Mathematical Olympiad, 1989)
[Solution] If $n$ is odd, then its factors $d_{1}, d_{2}, \cdots, d_{k}$ are all odd, and the sum of the squares of four odd numbers $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}$ is even and cannot equal the odd number $n$.
Therefore, $n$ is even.
Thus, $d_{1}=1, d_{2}=2$.
If $n$ is a multiple of 4, then
$$
4 \in\left\{d_{3}, d_{4}\right\} .
$$

In this case, among the squares of the first four factors $d_{1}, d_{2}, d_{3}, d_{4}$, there are already two ($2^{2}$ and $4^{2}$) that are multiples of 4, one is 1, and if another factor is odd, then its square is congruent to 1 modulo 4, at this time, $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}$ is a number of the form $4k+2$, and $n$ is not a multiple of 4; if another factor is even, then its square is divisible by 4, at this time $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}$ is a number of the form $4k+1$, and $n$ is also not a multiple of 4.
Therefore, $n$ is even, but not a multiple of 4.
Let $n=2m$, where $m$ is odd.
$d_{3}$ should be the smallest prime factor of $m$, so $d_{3}$ is odd, at this time, $d_{1}=1, d_{2}=2$, $d_{3}$ is odd, $d_{4}$ is even, thus by the problem, $d_{4}=2d_{3}$.

Therefore,
$$
\begin{aligned}
n & =1^{2}+2^{2}+d_{3}^{2}+4 d_{3}^{2} \\
& =5+5 d_{3}^{2} .
\end{aligned}
$$

Thus, 5 is a divisor of $n$.
Since $d_{4} \neq 4$, so $d_{4} \geqslant 6$.
Thus $d_{3}=5, d_{4}=10$. Then
$$
n=1^{2}+2^{2}+5^{2}+10^{2}=130 \text{. }
$$