There are $\tbinom{65}{2}=2080$ possible sums, so we can find two such that $a+b\equiv c+d\pmod{2016}$.

henderson wrote: $65$ distinct natural numbers not exceeding $2016$ are given. Prove that among these numbers we can find four $a,b,c,d$ such that $a+b-c-d$ is divisible by $2016.$ Do $a,b,c,d$ have to be different?

mjuk wrote: henderson wrote: $65$ distinct natural numbers not exceeding $2016$ are given. Prove that among these numbers we can find four $a,b,c,d$ such that $a+b-c-d$ is divisible by $2016.$ Do $a,b,c,d$ have to be different? yes, they are distinct numbers.

henderson wrote: $65$ distinct natural numbers not exceeding $2016$ are given. Prove that among these numbers we can find four $a,b,c,d$ such that $a+b-c-d$ is divisible by $2016.$ SOURCE: Azerbaijan Junior Mathematical Olympiad 2016 SOLUTION Note that we only care about the numbers' remainders when divided by $2016$, so we can replace the $65$ natural numbers with numbers in the range $0$ to $2016$. We can make the condition $a+b-c-d$ being divisible by $2016$ instead that $a+b-c-d = 0 \implies a+b=c+d$. Then, there are $2016 \cdot 2 = 4032$ possible sums (anything from $0$ to $4032$ inclusive). If two of the $65$ numbers have the same sum, then the problem is solved, so assume that no two of the $65$ numbers have the same sum. Then, there are $\binom{65}{2} = 4160$ possible sums. By the pigeonhole principle, we can find four $a,b,c,d$ such that $a+b-c-d$ is divisible by $2016$.

$\binom{65}{2} >2016$ This finishes the problem where $\binom{65}{2} $ equals to the number of all possible sums so we can always find such pairs