From IMO 2012 shortlist:
What is the maximum value of $k$ such that the set $S=\{1,2,\cdots,2018\}$ can be partitioned into $k$ disjoint pairs such that the sum of (the two numbers in) each pair is pairwise distinct and none of the sums exceeds $2018$.
I predict that the answer is $672$, that is when the pairs are $(1,2017),(2,2015),(3,2013),\cdots,(672,675)$. But I am not sure whether my prediction is right. Furthermore, I wanted to try proving that $673$ pairs is impossible, and I think some pigeonhole principle might be needed but I don't know how to make use of it. Any help is surely appreciated. Thanks!
N.B : If anyone got a solution without using the pigeonhole principle, that would be fine too!
Best Answer
This problem is in IMO 2012 shortlist, page 20. The answer is $\left\lfloor \frac{2n-1}{5}\right\rfloor$. In this case $807$.
Original post in the shortlist: