Assume the natural number $N$ has the $k \geq 4$ positive divisors $1 = d_1 < d_2 < \cdots < d_k=N$ s.t. $(1) \;\; d_1^2 + d_2^2 + d_3^2 + d_4^2 = N$. If $N$ is odd, then all of the divisors of $N$ are odd, yielding $N$ is even by equation (1). This contradiction means $N$ is even. Hence $d_2=2$. Moreover, if $4 | N$, then $4 \mid 1 + d_3^2 + d_4^2$ by equation (1), which is impossible. Hence we have $4 | N-2$. Clearly $N \neq 2$ by equation (1), which means $N$ has an odd divisor. Let $p$ be the smallest prime divisor of $N$. This assumption implies $d_3=p$ and $d_4=2p$. Summa summarum we have $(d_1,d_2 ,d_3,d_4) = (1,2,p,2p)$, which inserted in equation (1) result in $(2) \;\; 5(p^2 + 1) = N$. Consequently $5 | p$ by equation (2) (since $p|N$), yielding $p=5$, which inserted in equation (2) give us $N = 130$. Conclusion: The only solution of equation (1) is $ N=130$.

The only possibility is $N=130$, where we get $1^2+2^2+5^2+10^2=130$. If $N$ is odd, then all its factors are odd. Therefore the sum of the squares of the $4$ smallest positive divisors of $N$ will be even. Hence $N$ cannot be odd. Now consider when $N$ is even. Note that $1, 2$ are the smallest divisors of $N$. We want $N$ that can be written as $1^2+2^2+c^2+d^2=N$, where $c, d (c<d)$ are the third and fourth smallest divisors of $N$. Note that $c$ and $d$ must have different parities since $N$ is even. First consider when $c$ is even and $d$ is odd. It follows that $c=4$ and $d>4$. We have $1^2+2^2+4^2+d^2=21+d^2=N$, but the LHS is $2 \mod 4$, while $N$ is $0 \pmod 4,$ so this case is impossible. The final case is when $c$ is odd and $d$ is even. Since $c<d$, we must have $d=2c$ or $c=3, d=4$. In the case where $d=2c$, we have $1^2+2^2+c^2+(2c)^2=5(c^2+1)=N$. Since $5$ divides $N$, we must have $c=5$. This yields the answer of $N=130$. In the case where $c=3, d=4$, we have $1^2+2^2+3^2+4^2=30$, but $4$ does not divide $30$, so this case is not possible. Hence only $N=130$ works.

It's same as Balkan MO 1989 #1?????

link to BMO

same with BMO 1989#1