Is 1100 a valid state for this machine? A room starts out empty. Every hour, either 2 people enter or 4 people leave. In exactly a year, can there be exactly 1100 people in the room?
I think there can be because 1100 is even, but how do I prove/disprove it?

The process can be rephrased as making a required motion of $+2$ every time, with an optional additional motion of $-6$ at some steps.  The forward part of that motion will reach $+2N$ in $N$ steps, plus some backward motion divisible by $6$ to attain the final total motion of $T=+1100$.  Here $N$ and $T$ are given and the exact condition for a pair $(N,T)$ to be realizable is that $(2N - T)/6$ be a non-negative integer.  That integer is equal to the number of backward moves, whether thought of as $-4$'s or $-6$'s. 
This answer was found in reverse, by solving the linear $2 \times 2$ system for the number of $+2$ and $-4$ moves (given $N$ and $T$), and interpreting the condition for the numbers to be non-negative integers.
As explained in the previous answers, for the problem as posted, $2N = 2 \times 24 \times \text{(days in year)}$ is divisible by $6$ and $1100$ is not, so the conditions are not satisfied.
