# Pigeonhole principle nontrivial problem

if we have n + 2 natural numbers there are always 2 whose sum or difference is a multiple of 2n + 1. I tried to use the pigeonhole principle with remainder technique. If the numbers are divisible by 2n+1, then we have 2n possible remainders. However, the n+2 numbers will be matching 2n remainders and n+2<2n, so I am stuck. Any help is appreciated!

There are $2n+1$ residue classes $\pmod {2n+1}$. If two of your numbers are in the same residue class, then (by definition) their difference is divisible by $2n+1$, so let's suppose that they occupy $n+2$ distinct residue classes.
Now remark that we can divide the classes into pairs that add to a multiple of $2n+1$, namely $(1,2n),(2,2n-1),\cdots, (n,n+1)$ with $0$ left unpaired. We note that there are $n$ pairs. Even if $0$ is one of your choices, we would still have to make $n+1$ choices from those pairs. by the pigeon hole principle, therefore, if you have $n+2$ choices you must have chosen two from the same pair, and we are done.
Since you consider the sum and the difference on those natural numbers, it boils down to considering only the differnce on relative numbers (where each non-zero natural number has been duplicated as a negative one). Since $0$ could be among your $n+2$ numbers, by duplicating non-zero numbers, you obtain a set of at least $2n+3$ relative numbers. You want two of them to fall in the same residue class among $2n+1$. Pigeons cannot fit.
Note that with $n+1$ natural numbers, when the $0$ is used, you may obtain $2n+1$ relative numbers and they may live in pairwise distinct residue classes. Pigeons can sleep comfortably then.