, 2022

Solution:

We begin by observing that the number of possibilities for choosing a group of 6 favourite players is exactly
$$\binom{11}{6} = \frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 462$$
which is precisely half the number of fans. This implies that there have to be exactly 462 fans from Liechtenstein and 462 fans from Switzerland. Otherwise, one of these two countries would send at least 463 fans, but then, by pigeonhole principle, two fans from the same country would share the same 6 favourite players, which contradicts the problem statement.

We therefore know that there are 462 fans from each country and furthermore, for each of the possible $\binom{11}{6}$ combinations of favourite players, there are a Swiss fan and a Liechtensteiner fan who have exactly these 6 favourite players.

Now, if there was no player who gave an autograph to both a Swiss and a Liechtensteiner person, we could partition the players into two disjoint sets $A$ and $B$, such that only players from set $A$ gave their autograph to Swiss fans and only players from $B$ gave their autograph to Liechtensteiner fans. (If some player did not give an autograph to anyone, we can just put him in an arbitrary set). Since $A$ and $B$ are disjoint and their union contains 11 elements, one of the sets has at most 5 elements. Without loss of generality, let $|A| \leq 5$ and $|B| \geq 6$. However, by the observation in the previous paragraph, there must be a Swiss fan whose favourite players are all in the set $B$, so none of them gave any autographs to Swiss fans. This is a contradiction to the fact that everyone got an autograph from exactly one of their favourite players.

We conclude that there must have been a player who gave an autograph to at least one fan from Switzerland and at least one fan from Liechtenstein.