This problem might be related to some of the ones suggested above, I did not check carefully.

Assume that you have to prepare the calendar of a basketball league to which $2n$ teams take part, where $n$ is an integer $\geq 1$. The full season consists of $2n-1$ dates where in each date each of the team has to play against another one (different than itself :-)). This has to be done in such a way that, overall, every team meets every other team exactly once (we all know what the calendar of a basketball league look like).

The preparation of the calendar can easily be accomplished using some very very basic group theory as follow. Let $G$ be the group ${(\mathbf{Z}/2\mathbf{Z})}^n$ and define the set of dates of the calendar to be $G$ deprived by the identity element. Identify now the $2n$ teams of the league with $G$ by picking any bijection. Then the typical game of the calendar on the date $D$ will look like ($X$ vs $X+D$).

I learnt about the idea of using group theory to solve the "calendar problem" long time ago from a friend who attended the Scuola Normale Superiore in Pisa. I have the impression that the solution he explained to me then did not involve the particular group $G$ given above, but instead an arbitrary abelian group of order $2n-1$ parametrizing the dates of the calendar. Unfortunately that solution, more elegant but perhaps less practical (!), does not occur to me at the moment.

This problem might be related to some of the ones suggested above, I did not check carefully.

Assume that you have to prepare the calendar of a basketball league to which $2n$ teams take part, where $n$ is an integer $\geq 1$. The full season consists of $2n-1$ dates where in each date each of the team has to play against another one (different than itself :-)). This has to be done in such a way that, overall, every team meets every other team exactly once (we all know what the calendar of a basketball league look like).

A solution is given in the comment below. The solution I had first suggested was wrong.