Iranian Mathematical Olympiad

First we claim that for every positive integer $k$, the edges which their numbers are divisible by $k$ form a cluster. Indeed, suppose on the contrary that $S$ is the largest cluster such that the number of its edges is divisible by $k$ and $v\notin S$ is another such edge. Applying problem statement on edge $v$ and edges of cluster we get there exists a larger cluster with number on edges divisible by $k$.

Now suppose that $p\mid \left(\genfrac{}{}{0ex}{}{n}{2}\right)$ where $p$ is a prime number. Therefore the edges with numbers divisible by $\frac{1}{p}\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ form a cluster and the number of such edges is $p$. So

$$\begin{array}{rl}p& =\left(\begin{array}{c}t\\ 2\end{array}\right)\Rightarrow 2p=t(t-1)\\ & \Rightarrow \{\begin{array}{l}p\mid t\Rightarrow t-1\mid 2\Rightarrow t=2\text{ or }t=3\Rightarrow t=3,p=3\\ p\mid t-1\Rightarrow t\mid 2\Rightarrow t=1\text{ or }t=2\Rightarrow \text{Contradiction!}\end{array}\end{array}$$

Thereby the only prime divisor of is three, so where $a\in \mathbb{N}$. If $a=1$ and so $n=3$ and numbers $1,2,3$ satisfies problem statement. If $a>1$ there exist $9$ multiple of $\frac{1}{9}$ among numbers, but $9$ cannot be written in the form so for $n>3$ such numbers do not exist. $\square$