This question already has an answer here:
In a complete graph with $n$ vertices, every vertex has $n-1$ edges. Assuming $n$ is odd, the number of edges from each vertex is even. If we now give every edge an orientation (making the graph a tournament), is it always possible to arrange the orientations such that every vertex has the same number of incoming and outgoing edges (apparently called a balanced directed graph)?
It is easy to give edges an orientation where the above is not true. E.g. orient all edges to a given vertex as incoming edges:
But it is also relatively easy to orient the edges so the above is true, at least for $n=3, 5, 7, 9$, which is what I've tried so far. An example for $n=5$ is given below:
But is it always possible for any odd $n$?