passing cards around a table $71$ people are seated around a circular table. Each of the numbers $1, 2, 3, . . . , 71$ is written on two pieces of yellow paper and, randomly, each person is given two of the pieces of yellow paper. Once a minute each person passes to the person to their right the piece of yellow paper with the smaller of their two numbers. Prove/disprove that there must come a time when someone’s two pieces of yellow paper contain the same number.
So I have found a counterexample when $n$ is $70$: give every person one piece of paper containing $1$-$35$ and every person one piece of paper containing $36$-$70$. Then the $1$-$35$ will always be passed and they will never intersect. However with $n=71$ this becomes really strange: I believe that there still exist a counterexample but I cannot find it out. Some suggestion or help will be appreciated.
 A: Let $n$ be the number of participants, and suppose $n$ is odd (e.g., $n=71$).

If $n=1$, the truth of the claim is obvious.

Thus, assume $n \ge 3$.

Label the participants, in counterclockwise order, as $P_1,...,P_n$, and for convenience, let $P_0$ be an alternate notation for $P_n$.

Let $t$ represent "time", measured as the number of transitions which have transpired.

For each nonnegative integer $t$, let 
$$s(t) = \Bigl(\bigl(x_1(t),y_1(t)\bigr),...,\bigl(x_n(t),y_n(t)\bigr)\Bigr)$$
be the state at time $t$, where $x_k(t),y_k(t)$ are the min and max values for $P_k$ at time $t$.

Since the number of possible states is finite, the states must eventually repeat in a cycle of length $m$, say.

In every transition, of the two values of $1$, at least one of them must move so as to replace a value greater than $1$, hence a transition always yields a state different than the previous one.

It follows that $m > 1$.

To prove the claim, it suffices to prove it assuming the cycle starts at time $t=0$.

Thus, we have states
$$s(0),...,s(m)$$
with $s(0),...,s(m-1)$ distinct, and $s(m)=s(0)$.

Let $X(t),Y(t)$ be defined by
\begin{align*}
X(t) &= x_1(t) + \cdots + x_n(t)\\[4pt]
Y(t) &= y_1(t) + \cdots + y_n(t)\\[4pt]
\end{align*}
Note that in any transition from time $t$ to time $t+1$, either $y_k(t+1) = y_k(t)$, or if not, then $x_k(t+1) = y_k(t)$ and $y_k(t+1) > y_k(t)$.

It follows that
$$Y(0) \le Y(1) \le \cdots \le Y(m)$$
but then, since the states cycle at time $t=m$, it follows that
$$Y(0) = Y(1) = \cdots = Y(m)$$
and hence, we also have
$$X(0) = X(1) = \cdots = X(m)$$
Then for each $k\in \{1,...,n\}$, we must have
$$y_k(0) = y_k(1) \cdots = y_k(m)$$
and also, for each positive integer $t$, we have
$$x_k(t) = x_{k-1}(t-1)$$
In other words, in every transition, the high values don't move, and the low values move right by $1$.

It follows that the highest low value $L$ say, is less than or equal to the lowest high value, $H$ say.

Thus, $L \le H$.

Suppose $L < H$.

Then all the low values are less than all the high values.

But each of the low values occurs with multiplicity $2$, contradiction, since the sequence of low values has length $n$, which is odd.

Then we must have $L=H$.

Since $L$ is a low value, it always moves right, and since $H$ is a high value, it always stays put.

It follows that the value $L$ will eventually be the low value for a participant whose high value is $H$.

This completes the proof.
