There are three solutions: $n = 6$, $n = 8$ and $n = 12$.

First, notice that if $n$ is odd, all its divisors are also odd, so if $d_1$ and $d_2$ are two divisors such that $1 < d_1 < d_2 < n$, then $d_2 - d_1$ is even, so $d_2 - d_1 \nmid n$. Therefore, $n$ is even.

Consider $n = 2^a \cdot b$, where $a, b \in \mathbb{N}_{\ge 1}$ and $b$ is odd.
If $b = 1$, then $n = 2^a$, with $a \in \mathbb{N}^*$. For $a \in \{1, 2\}$ there are no solutions, and for $a \ge 4$, choosing $d_1 = 2^{a-3}$ and $d_2 = 2^{a-1}$, it follows that $d_2 - d_1 = 3 \cdot 2^{a-3} \nmid n$, a contradiction. Consequently, $a = 3$, so $n = 8$. It's trivial to see that $n = 8$ is a solution of the problem.

For $b \ge 3$ we have two cases.
If $a = 1$, then $n = 2b$. Choose $d_1 = 2$ and $d_2 = b$; we obtain $b - 2 \mid 2b$, so $b - 2 \mid 2b - 2(b - 2) = 4$. Since $b$ is odd, we infer that $b = 3$, so $n = 6$, which is a solution.

If $a \ge 2$, choose $d_1 = 2^a$ and $d_2 = 2^{a-1} \cdot b$; we obtain $2^{a-1}(b-2) \mid 2^a \cdot b$, so $b - 2 \mid 2b$. Since $b$ is odd, similarly it follows that $b = 3$. Therefore, $n = 2^a \cdot 3$.
Choose now $d_1 = 3$ and $d_2 = 2^a$; it results that $2^a - 3 \mid 2^a \cdot 3$, so $2^a - 3 \mid 2^a \cdot 3 - 3 \cdot (2^a - 3) = 9$, from which we obtain $a = 2$. Hence, $n = 12$ and we easily verify that it is a solution.