Determine all natural numbers $n$ with two different positive divisors that are equally distant from $\frac{n}{3}$.

(Richard Henner)
0 is of course not a solution, because all different natural numbers have different distances from 0.

Let $s$ and $t$ be the two divisors that have the same distance from $\frac{n}{3}$, and w.l.o.g. $s<\frac{n}{3}<t$. If $s$ and $t$ have the same distance from $\frac{n}{3}$, then $\frac{s+t}{2}=\frac{n}{3}$, so $3(s+t)=2 n$. It follows that $n$ must be divisible by 3.

Now let $n=3 k$ and $k$ be an odd number. Then 3 is the smallest divisor of $n$ greater than 1, and thus $\frac{n}{3}$ is the largest divisor of $n$ less than $n$. Then $t=n$ and $s=-\frac{n}{3}$, so $s$ is not positive. Therefore, an odd number cannot be a solution.

Let $n=6 k$. Then $\frac{n}{3}=2 k$ and $s=k$ and $t=3 k$ are divisors of $6 k$ and have the same distance from $2 k$.

Therefore, all positive multiples of 6 are solutions.

(Richard Henner)