55rd Ukrainian National Mathematical Olympiad - Third Round

Answer: $n\in \{1;2;3;4;6\}$.

Clearly a number cannot have divisors greater than $\frac{n}{2}$, besides $n$ itself. Therefore, in order to have more than $\frac{n}{2}$ divisors it must be divisible by all numbers from $1$ to $\frac{n}{2}$ and by $n$. Denote by $m$ the integer number which equals either $\frac{n}{2}$ or $\frac{n-1}{2}$. Then if $n\ge 10$, either $m$ or $m-1$ is coprime with $3$. Hence, $n\ge 3m$, because it must be divisible by both $3$ and $m$. However, in this case $n\ge 3\cdot (\frac{n-1}{2}-1)$, or $2n\ge 3n-9$, which contradicts the assumption that $n\ge 10$. All numbers $n<10$ can be checked by hand.