A number theory question For integers from $1$ to $4n-1$, we pick two integers and replace them with their difference. That is for $n=1$ we have $1,2,3$ if we pick $1,3$ our new numbers will be $2,2$. Show that at $4n-2$ step the remaining number would be even. This is a question from a local mathematical contest, I couldn't figure out the solution. Thanks for any help!
 A: We rephrase the problem in order to make it more explicit. At the beginning, we have the numbers $1$ to $4n-1$ written on a blackboard. At every step, we take two of the numbers on the blackboard, say $a$ and $b$, erase them, and write $a-b$, or perhaps $|a-b|$, on the blackboard. So after $4n-2$ steps, there will be only $1$ number on the blackboard. Show this number must be even.

Consider the parity of the sum. At the beginning, the sum of the numbers on the board is $1+2+\cdots+(4n-1)$. This is even. That can be seen by noting that the sum is $\frac{(4n-1)(4n)}{2}$, and in other ways.  
When we replace the two numbers $a$ and $b$ by the single number $a-b$, or $|a-b|$, the parity of the sum does not change. This is because $a-b$ has the same parity as $a+b$. (They are both even or both odd.) So since the sum started out even, the sum remains even. That means that when we are down to a single number, that number must be even.
A: Hint $\ $ The operation preserves the fact that the list contains an even number of odd integers. Therefore, when you reach a list of one element $\,k,\,$ the list has has zero odds, so $\,k\,$ is even. 
