Since $n$ is a composite number, we have $k \ge 3$.
Let $d_2 = p$ be the smallest prime that divides $n$. We show by induction that
$$d_j = \frac{j(j-1)}{2}p - \frac{(j-2)(j+1)}{2}, \quad j = 1, 2, \dots, k.$$

This is clearly true for $j = 1$ and the induction step follows from $d_j - d_{j-1} = (j-1)(d_2 - d_1) = (j-1)(p-1)$ and $1 + 2 + 3 + \dots + (j-1) = \frac{j(j-1)}{2}$.
If we apply this formula to $d_{k-1} = \frac{n}{p} = \frac{d_k}{d_2}$ and multiply by $2p$, we get
$$\begin{aligned} & (k-1)(k-2)p^2 - (k-3)kp = k(k-1)p - (k-2)(k+1) \\ \Leftrightarrow \quad & (k-1)(k-2)p^2 - 2(k-2)kp + (k-2)(k+1) = 0 \\ \Leftrightarrow \quad & (k-1)p^2 - 2kp + (k+1) = 0. \end{aligned}$$
The solutions of this quadratic equation are $p = 1$ and $p = \frac{k+1}{k-1} = 1 + \frac{2}{k-1}$. Since both options are at most 2, the only possibility is $p = 2$, $k = 3$ and $n = 4$. Since $n = 4$ has the required property, this is the only solution.