IRL_ABooklet

There are $4\times 4=16$ possible combinations of morning and evening trains. As $2022>2016=16\times 126$ there must be one combination of morning and evening trains with at least $127$ Abelians on both trains. Identify these Abelians, for example by painting them purple.

Suppose for a contradiction that no two purple Abelians live in houses whose numbers differ by $5$ or less. Then ordering the Abelians by their house numbers, the first purple Abelian is in house number at least $1$. There can be no purple Abelians in houses $2,3,4,5$ or $6$. The second purple Abelian must then be in house number at least $7$, until the $127$th purple Abelian must be in house number at least $757=1+6\times 126$.

As the average household size is $3$, there are only $674=2022/3$ houses in the village, which is a contradiction.

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Alternative solution.

Because the average household size is $3$, there are $674=2022/3$ houses in the village. These houses can be grouped into $113$ groups such that the house numbers differ by at most $5$ in each group, for example, for $k=1,2,\dots ,112$ the six houses $6k-5,\dots ,6k$ form one group, and the remaining two houses $673,674$ form the last group.

Because $16\cdot 113=1808<2022$, there must be at least one group of houses in which at least $17$ people live. There are $4\times 4=16$ possible combinations of morning and evening trains, hence there are two among these $17$ Abelians who were on the same train both in the morning and in the evening.