The theorem I am trying to prove says:
Given $n+1$ natural numbers, say $a_1, a_2,...,a_{n+1}$, all less than or equal to $2n$, then there exists a pair, say $a_i$ and $a_j$ with $i \neq j$ , such that $a_i\,|\,a_j$.
I tried induction, but its not feasible, so I came up with a pigeonhole proof. I am not sure if my proof is complete so I still require some help.
Proof: Consider all pairs of numbers of the form $(k,2k)$ with $k$ between $1$ and $n$. If $n$ is $10$, consider the following pairs: $(1,2)\\(2,4)\\(3,6)\\(4,8)\\(5,10)$
Our list contains $n+1$ numbers. But if we choose two numbers from any of these $n$ pairs, one will divide the other. By the pigeonhole principle we must pick two numbers from one of these pairs. $\blacksquare$