Let $d_1$, $d_2$, $d_3$, $d_4$ be the smallest four positive divisors of $n$. We have $d_1 = 1$ and $$n-1={d_2}^2+{d_3}^2+{d_4}^2$$If $n$ were odd, then $d_1$, $d_2$, $d_3$, $d_4$ are all odd and $n$ is even, contradiction. So $n$ is even, therefore $d_2=2$. The equation reduces to $$n-5={d_3}^2+{d_4}^2$$. From here, ${d_3}^2+{d_4}^2$ must be odd, since $n$ is even. Thus one of $d_3$ and $d_4$ is odd while the other is even. Case 1: If $d_4$ is even, say $d_4=2m$. If $m=2$, $d_3$ must be $3$ and $n=1^2+2^2+3^2+4^2=30$, but $d_4=4$ does not divide $30$, so $n=30$ is not a solution. If $m>2$, since $m$ must be a divisor (as $2m$ is a divisor), $d_3=m$ and $n=5+5m^2$. Finally, as $m \mid n$, $$m \mid 5+5m^2 \implies m \mid 5 \implies m=5$$and $n=130$, which satisfies the problem's conditions. Case 2: If $d_3$ is even, and there were some $m \geq 3$ such that $d_3=2m$, then $m$ would be a divisor between $d_2$ and $d_3$, contradiction. So $m \leq 2$, and since $m=1$ yields $d_3=2$ which is absurd, $m=2$ and $d_3=4$. Hence, if we say $d_4=k$, $n=1^2+2^2+4^2+k^2=21+k^2$, and from $k \mid n$ we get $k \mid 21$, so $k=7$. However, this yields $n=70$ which does not satisfy the conditions since $4$ does not divide $70$. Thus, the only $n$ that satisfies the problem's conditions is $130$.

Let \( n_i \) be the \( i \)-th factor of \( n \). \[ n = 1 + (n_2)^2 + (n_3)^2 + (n_4)^2 \] Case 1: \( n \) is even If \( n \) is even, \( n_2 = 2 \). Case 2: \( n \) is odd If \( n \) is odd, \( n_2, n_3, n_4 \) are all odd. This means \( 1 + (n_2)^2 + (n_3)^2 + (n_4)^2 \) will be even, causing a contradiction. Thus, \( n \) is even, and \( n_2 = 2 \). \[ n = 5 + (n_3)^2 + (n_4)^2 \] We can see that either \( n_3 \) or \( n_4 \) is even. Subcase 1: \( n_3 \) is even If \( n_3 \) is even, then \( \frac{n_3}{2} \) has to be a smaller factor of \( n \). Thus, if \( n_3 \) is even, \( n_3 = 4 \). \[ n = 21 + (n_4)^2 \] Let \( n = a n_4 \) where \( a > 1 \). \[ a n_4 - (n_4)^2 = 21 \] \[ a - n_4 = \frac{21}{n_4} \] \( n_4 \) is an odd factor of 21 greater than 4, so \( n_4 = 7 \). \[ n = 21 + 49 = 70 \] However, the 4 smallest factors of 70 are \( 1, 2, 5, 7 \). Thus, by contradiction, \( n_3 \) is odd, and \( n_4 \) is even. Subcase 2: \( n_4 \) is even Let \( n_4 = b n_3 \) where \( b > 1 \) and \( b \) is not a multiple of 4. \[ n = 5 + (b+1)(n_3)^2 \] Since \( n \) and \( (b+1)(n_3)^2\) are divisible by \( n_3 \), thus \(5 \) is divisible by \( n_3 \). This must mean \( n_3 = 5 \). If \( n_4 = 15 \), then \( n_3 = 3 \), which is a contradiction. Thus, \( n_4 = 10 \). \[ n = 1^2 + 2^2 + 5^2 + 10^2 = 130 \] Final Answer: \[ n = 130 \]