National Competition

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 $$d_j=\frac{j(j-1)}{2}p-\frac{(j-2)(j+1)}{2},j=1,2,\dots ,k.$$

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 $1+2+3+\cdots +(j-1)=\frac{j(j-1)}{2}$. If we apply this formula to $d_{k-1}=\frac{n}{p}=\frac{d_k}{d_2}$ and multiply by $2p$, we get $$\begin{array}{rl}& (k-1)(k-2)p^2-(k-3)kp=k(k-1)p-(k-2)(k+1)\\ \iff & (k-1)(k-2)p^2-2(k-2)kp+(k-2)(k+1)=0\\ \iff & (k-1)p^2-2kp+(k+1)=0.\end{array}$$ 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.