Determine all composite positive integers $n$ with the following property: If $1=d_{1}<d_{2}<\ldots<d_{k}=n$ are all positive divisors of $n$, then

$$
\left(d_{2}-d_{1}\right):\left(d_{3}-d_{2}\right): \cdots:\left(d_{k}-d_{k-1}\right)=1: 2: \cdots:(k-1) .
$$

(Walther Janous)
Since the number $n$ is composite, we have $k \geq 3$. With $d_{2}=p, p$ prime, it follows from $(p-1):\left(d_{3}-p\right)=1: 2$ that $d_{3}=3 p-2$. From $(p-1):\left(d_{4}-d_{3}\right)=1: 3$ we get $d_{4}=6 p-5$. By induction, one can see that $d_{j}=(1+2+\ldots+(j-1)) p-(2+3+\ldots+(j-1))$ holds, thus

$$
d_{j}=\frac{j(j-1)}{2} p-\frac{(j-2)(j+1)}{2}, \quad j=1,2, \ldots, k .
$$

In particular, since $d_{k-1}=\frac{n}{p}=\frac{d_{k}}{d_{2}}$, we have

$$
(k-1)(k-2) p^{2}-(k-3) k p=k(k-1) p-(k-2)(k+1)
$$

which must hold, thus

$$
(k-1)(k-2) p^{2}-2(k-2) k p+(k-2)(k+1)=0
$$

Therefore, the equation

$$
(k-1) p^{2}-2 k p+(k+1)=0
$$

remains to be considered. It has $p=1$ or $p=\frac{k+1}{k-1}=1+\frac{2}{k-1}$ as its solutions. Since both options for the prime $p$ can take a maximum value of 2, the only possible values are $k=3$ with $p=2$ and $n=4$. Since the condition for $n=4$ is obviously satisfied, this is the only solution.

(Walther Janous)