Blackboard game

The sum of these $110$ numbers, $6105$ , is odd. If we wipe out the "random" numbers $a$ and $b$ and write their difference $a-b$ instead, then the sum of the remaining 109 numbers, $6105-2b$ , will again be odd. Likewise, we will end up with an odd sum in each consecutive step.

You start with 55 odd numbers and 55 even numbers. On any given step you will either:

1) Pick two odds (decreasing the number of odds by two) and replace their even difference (increasing the number of evens by one);

2) Pick two evens (which will erase two evens, and replace one, or a net decrease of evens by one); or

3) Pick one even and one odd, with a difference that is odd (odds stay the same, and evens decrease by one).

So, while the even numbers can either go up or down by one, the odds will either stay the same or decrease by two. Both groups will ultimately decrease (as the net numbers go down each time). But as the number of odd numbers will always be odd, eventually you will have one number left, which must be odd.