1. Let $d_1, d_2, d_3, d_4$ be the smallest four positive divisors of $n$. We know that $d_1 = 1$.
2. Given that $n$ is the sum of the squares of its four smallest factors, we have:
$$n = d_1^2 + d_2^2 + d_3^2 + d_4^2$$
Substituting $d_1 = 1$, we get:
$$n = 1^2 + d_2^2 + d_3^2 + d_4^2 = 1 + d_2^2 + d_3^2 + d_4^2$$
Therefore:
$$n - 1 = d_2^2 + d_3^2 + d_4^2$$
3. If $n$ were odd, then $d_1, d_2, d_3, d_4$ would all be odd, making $n$ even, which is a contradiction. Hence, $n$ must be even, implying $d_2 = 2$.
4. Substituting $d_2 = 2$ into the equation, we get:
$$n - 5 = d_3^2 + d_4^2$$
5. Since $n$ is even, $d_3^2 + d_4^2$ must be odd. This means 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 $m = 2$, then $d_3$ must be 3, and:
$$n = 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30$$
However, $d_4 = 4$ 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:
$$n = 5 + 5m^2$$
Since $m \mid n$:
$$m \mid 5 + 5m^2 \implies m \mid 5 \implies m = 5$$
Therefore:
$$n = 5 + 5 \cdot 5^2 = 5 + 125 = 130$$
We need to check if 130 satisfies the problem's conditions. The divisors of 130 are 1, 2, 5, 10, 13, 26, 65, and 130. The smallest four divisors are 1, 2, 5, and 10. We check:
$$1^2 + 2^2 + 5^2 + 10^2 = 1 + 4 + 25 + 100 = 130$$
Thus, $n = 130$ satisfies the 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 $d_3$, which is a contradiction. So $m \leq 2$.
- Since $m = 1$ yields $d_3 = 2$, which is absurd, $m = 2$ and $d_3 = 4$. Hence, if we say $d_4 = k$:
$$n = 1^2 + 2^2 + 4^2 + k^2 = 1 + 4 + 16 + k^2 = 21 + k^2$$
From $k \mid n$, we get $k \mid 21$, so $k = 7$. However, this yields:
$$n = 21 + 7^2 = 21 + 49 = 70$$
This does not satisfy the conditions since 4 does not divide 70.

Thus, the only $n$ that satisfies the problem's conditions is $130$.

The final answer is $\boxed{130}$.