I'm struggling solving this problem : I need to prove that given any set of 1011 positive integers, there are 2 numbers such that $x_1 + x_2$ or $x_1 - x_2$ is a multiple of 2020.
basically I understand that I need to show that there are 2 numbers such that $ x_1 + x_2 = 0 \pmod{ 2020} $ or $ x_1 - x_2 = 0\pmod {2020}$ but can't find a way to solve that.
Hope to get a help.
edit - also able to make $x_1+x_1=0 (mod 2020)$ if necessary
As stated, the problem is flawed; $1011$ numbers is not enough.
For example, if $x,y$ are two distinct elements of the $1011$-element set $ S=\{0,...,1010\}$, then $$ \left\lbrace \begin{align*} &x+y\in\{1,...,2019\}\\[4pt] &|x-y|\in\{1,...,1010\}\\[4pt] \end{align*} \right. $$ hence neither of $x+y,x-y$ is divisible by $2020$.
But $1012$ numbers is enough.
Claim:
If $S$ is a set of integers with $|S|\ge1012$, then there exist two distinct elements of $S$ whose sum or difference is divisible by $2020$.
Proof:
If two distinct elements of $S$ are congruent to each other mod $2020$, then their difference is a multiple of $2020$, and we're done.
Thus assume no two distinct elements of $S$ are congruent to each other mod $2020$.
Then without loss of generality, we can reduce the elements of $S$ mod $2020$, hence we can assume $S\subseteq\{0,...,2019\}$.
Our goal now is to show that there are two distinct elements of $S$ whose sum is $2020$.
Let $A=S{\setminus}\{0,1010\}$ and let $B=\{2020-a{\,\mid\,}a\in A\}$.
Then $A,B$ are subsets of the $2018$-element set $$ \{1,...,1009\}\cup\{1011,...,2019\} $$ hence, since $|A|\ge 1010$ and $|B|=|A|$, it follows by the pigeonhole principle that $A\cap B\ne {\large{\varnothing}}$.
Let $x\in A\cap B$.
Since $x\in B$ we have $x=2020-y\;$for some $y\in A$.
Then $x,y\in A$ and $x+y=2020$.
If $x=y$ then $x+y=2020$ would imply $x=1010$ and $y=1010$, contradiction, since $1010\not\in A$.
Thus $x,y$ are two distinct elements of $S$ whose sum is $2020$.
This completes the proof,