Beginners' Competition

Since the smallest possible divisors of an integer $n$ are $1$, $2$ and $3$, the greatest possible divisors are $n$, $\frac{n}{2}$ and $\frac{n}{2}$. Hence a divisor that is bigger than $\frac{n}{2}$ can only be $n$ or $\frac{n}{2}$. Since there is no positive divisor of $n$ having the same distance from $\frac{n}{2}$ as $n$, the bigger one of the two divisors must be $\frac{n}{2}$. The distance from $\frac{n}{2}$ to $\frac{n}{2}$ equals $\frac{n}{2}$ and since $\frac{n}{2}-\frac{n}{2}=\frac{n}{2}$ the smaller divisor must be $\frac{n}{2}$. Hence $n$ is a multiple of $6$. On the other hand it is clear that all positive multiples of $6$ have the desired property.