Here's the problem:
There are 25 people sitting around a table and each person has two cards. One of the numbers 1,2,..., 25 is written on each card, and each number occurs on exactly two cards. At a signal, each person passes one of her cards, the one with the smaller number to her right hand neighbor. Prove that sooner or later, one of the players will have two cards with the same numbers.
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In the second solution, why would $A_{max}$ remain unchanged after 25 moves? And why is $a_{min}(i) < a_{max}(j)$ for all i,j if $A_{max}$ is unchanged? I was thinking that if not, then we could find some $i,j$ so that $a_{min}(i) \ge a_{max}(j)$ and then on the next move, there'd be a change in $A_{max}$. But I can't see why there's a contradiction.
In fact, $A_\text{max}$ will eventually remain unchained for infinitively many moves. If $A_\text{max}$ was not eventually constant, then it would have to increase infinitely many times, as it can never decrease. This is impossible, since $A_\text{max}\le 25^2$ always.
First of all, notice that if person $i$ receives a number which is greater than $a_\text{max}(i)$, then $a_\text{max}(i)$ increases, so $A_\text{max}$ increases. Contrapositively, since $A_\text{max}$ does not increase, over the next $25$ turns, all of the objects which are passed must be less than $a_\text{max}(i)$, for all $i\in \{1,\dots,25\}$. The items which are being passed are precisely the items $a_\text{min}(j)$, for $j\in \{1,\dots,25\}$.