TITLE: Can we find two numbers from $n+2$ numbers chosen from $\{1,2,3,\cdots\}$?
QUESTION [5 upvotes]: Let $n\ge 2$ is given and choose $n+2$ integers from the set $\{1,2,3,\cdots,3n\}$ then can we find two numbers whose difference is greater than $n$ but less than $2n$?
I try to prove it by using pigeon principle. I partitioned $\{1,2,3,\cdots,3n\}$ from the sets
$$\{\{k,2n-1+k\}:1\le k\le n+1\}\cup \{\{k\} :n+2\le k\le 2n-1\}.$$
However I don't know how to proceed it. Thanks for any help. (It is homework problem so I don't want a full solution, but a hint or idea.)

REPLY [3 votes]: Consider a set of $n+1$ numbers. The first number in the list removes $n-1$ candidates - i.e. choosing any one of these results in a success. Every other choice from the list removes at least one more candidate, and there are $n$ of them. So we have the $n+1$ chosens and $n-1+n$ removals, $=3n$ in total, so there is nowhere left to go.