The real point is that there will always be an odd number of red balls. We start with an odd number and there are only 3 operations:

Remove 2 reds - add a blue - keeps red odd

Remove two blues - add a blue - keeps red odd

Remove one of each - add a red - keeps red odd.

Then whenever you get down to one red you can't get rid of it, since you will be always removing one red and one blue, then adding back a red.

To answer this, assume that you always perform the same action each time (e.g remove 2 red, remove 2 blue, remove R and B) and see what you end up with. The answer is you will always end up with 1 red 1 blue in the pot, so you remove these and replace with a red. So the final ball removed is always red.