2021 IMO Shortlist C4

The kingdom of Anisotropy consists of $n$ cities. For every two cities there exists exactly one direct one-way road between them. We say that a path from $X$ to $Y$ is a sequence of roads such that one can move from $X$ to $Y$ along this sequence without returning to an already visited city. A collection of paths is called diverse if no road belongs to two or more paths in the collection. Let $A$ and $B$ be two distinct cities in Anisotropy. Let $N_{A B}$ denote the maximal number of paths in a diverse collection of paths from $A$ to $B$. Similarly, let $N_{B A}$ denote the maximal number of paths in a diverse collection of paths from $B$ to $A$. Prove that the equality $N_{A B}=N_{B A}$ holds if and only if the number of roads going out from $A$ is the same as the number of roads going out from $B$.

各向异性王国由 $n$ 个城市组成。对于每两个城市，它们之间恰好存在一条直接的单向道路。我们说从 $X$ 到 $Y$ 的路径是一系列道路，这样人们可以沿着这个序列从 $X$ 移动到 $Y$ 而无需返回已经访问过的城市。如果没有一条道路属于集合中的两条或更多条路径，则该路径的集合称为多样化的。令 $A$ 和 $B$ 为 Anisotropy 中两个不同的城市。令 $N_{A B}$ 表示从 $A$ 到 $B$ 的不同路径集合中的最大路径数。类似地，令 $N_{B A}$ 表示从 $B$ 到 $A$ 的不同路径集合中的最大路径数。证明等式$N_{A B}=N_{B A}$成立当且仅当从$A$出去的道路数量与从$B$出去的道路数量相同。