Combinatorics Problem: Box Riddle A huge group of people live a bizarre box based existence. Every day, everyone changes the box that they're in, and every day they share their box with exactly one person, and never share a box with the same person twice.
One of the people of the boxes gets sick. The illness is spread by co-box-itation. What is the minimum number of people who are ill on day n?

Additional information (not originally included in problem):
Potentially relevant OEIS sequence: http://oeis.org/A007843

If we want to minimize the number of infected people only after a fixed number $n$ of days -- meaning we do not an answer that is a function of $n$ as $n$ tends to infinity -- then the answer depends on the parity of $n$: for odd $n$, the answer is $n + 1$, while for even $n$ the answer is $n + 2$.
First observe that there cannot be less then $n + 1$ infected individuals at day $n$ since the first infected guy has shared boxes with at least $n$ other indiviuals.
Let us now assume $n$ is odd. Take a set $V$ containing $n + 1$ citizens of the box world (one of them is the infected one). Take a complete graph on $V$. Decompose its edges into perfect matchings $M_1, \dots, M_{n}$. Each perfect matching $M_i$ tells you how the co-box-itation will be at day $i$. Assume people outside of $V$ share boxes only with people outside of $V$ during these $n$ days. This scheme makes everyone in $V$ sick at day $n$ while nobody outside of $V$ is sick at day $n$. 
For $n$ days, with $n$ even, the proof that there can be as little as $n + 2$ infected individuals is similar to the previous paragraph. 
The proof that there cannot be less than $n + 2$ infected individuals is as follows. We already know that, at the end of day $n - 1$, there are at least $n$ infected individuals (because $n - 1$ is odd). If there are precisely $n$, it is because they have shared boxes among themselves during these $n - 1$ days. Since they are not allowed repeat partners, on the $n$-th day, each has to share a box with a non infected individual, which leads to $2n$ infected individuals at the end of day $n$. If there are exactly $n + 1$ infected individuals at the end of day $n - 1$, then it is still possible that some of them share a box on day $n$, but $n$ is even, so one of these $n + 1$ individuals will be forced to share a box with a healthy person.
-- EDITED --
In fact, every complete graph of even order may be decomposed into perfect matchings (see Theorem 3.5 on page 20 of the book One-Factorizations by W. D. Wallis). 
