75th Romanian Mathematical Olympiad

Answer: all primes and $n=6$. The primes have the required property, because the divisors of a prime $n$ are $1=d_1<d_2=n$, and $n$ is divisible by $d_1$.

Take now a positive integer $n$ with $k\ge 3$ divisors and fulfilling the required property. If $n$ is odd, then all its $k$ divisors are odd. From the condition $n$ must be divisible by $d_1+d_2$, which is even – a contradiction. Therefore, $n$ must be even.

Then $d_1=1$, $d_2=2$, so $n$ must be divisible by $d_1+d_2=3$, hence $d_3=3$. Moreover, $6\mid n$. This shows that $\frac{n}{6}$, $\frac{n}{3}$ and $\frac{n}{2}$ are divisors of $n$.

Since $n$ is divisible by $d_1+d_2+\cdots +d_{k-1}$, we get $n\ge d_1+d_2+\cdots +d_{k-1}\ge \frac{n}{6}+\frac{n}{3}+\frac{n}{2}=n$. This shows that the only divisors of $n$, smaller than $n$, are $\frac{n}{6}$, $\frac{n}{3}$ and $\frac{n}{2}$, hence $\frac{n}{6}=d_1=1$, $\frac{n}{3}=d_2=2$ and $\frac{n}{2}=d_3=3$, yielding $n=6$, which fulfills the condition.