IMO 2020 Problem 3

This is the worst combinatorial problem I've seen in IMO. Well, yes I'm biased by my failure to make progress until getting hint, something too advanced and specialized but also makes the problem very simple once you get it. There doesn't seem to be any other approach to it. Moreover rumor has it that some students were able to draw inspiration from here.

Problem:

There are $4n$ pebbles of weights $1, 2, 3, \dots, 4n.$ Each pebble is colored in one of $n$ colors and there are four pebbles of each color. Show that we can arrange the pebbles into two piles so that the following two conditions are both satisfied:

-The total weights of both piles are the same.

-Each pile contains two pebbles of each color. 

===================================

Proof:

The strategy is to pair $k$ with $4n+1-k$. In a graph $G$ with $n$ vertices each representing a color, two colors $a$ and $b$ have a connecting edge for every pair of pebbles of colors $a$ and $b$ that sum up to $4n+1$. Note that multiple edges can exist between a pair of colors or even a self loop. So $G$ is $4$-regular and therefore has a $2$-factorization, and the partitioning follows immediately which clearly satisfies all conditions.

Q.E.D.