We say the integers $(a_1, a_2, \ldots, a_{2m})$ form an ordered integer cycle if the following holds: Per writes $a_1$ on the same card as $a_2$; Kari writes $a_2$ on the same card as $a_3$; $\vdots$ Kari writes $a_{2m}$ on the same card as $a_1$. Clearly, each integer between $1$ and $2n$ inclusive belongs to a unique ordered integer cycle containing at least two elements. In addition, it's easy to see that all the integers in an ordered integer cycle can be shown (face up) exactly once. These two observations imply the desired result. $\blacksquare$ Remark: This construction and solution can be motivated by the following process: Assume Per has a card with $a_1$ and $a_2$ written on it. Suppose they choose to have $a_1$ facing up. This forces Keri's card with $a_2$ written on it to have $a_2$ facing up. If the other integer on Keri's card is $a_3$, then Per must force their card with $a_3$ written on it to have $a_3$ facing up. Now, one can reasonably guess that this process terminates at a certain point. The main claim follows naturally from this observation. Note: The process of choosing which face to show feels somewhat symmetrical, so assuming $a_1$ is facing up isn't a huge leap of faith.

parmenides51 wrote: Per and Kari each have $n$ pieces of paper. They both write down the numbers from $1$ to $2n$ in an arbitrary order, one number on each side. Afterwards, they place the pieces of paper on a table showing one side. Prove that they can always place them so that all the numbers from $1$ to $2n$ are visible at once. Consider a directed graph $G$ such that $a_1 \to a_2$ if $a_1$ and $a_2$ are on the same card and $a_1$ is shown. If Peri chooses $G$ then Keri reverses all arrows in Peri's graph which contains all of $1,2, \dots 2n$

trival by graph