Parity of red balls decides the final colour

A bag starts with $20$ blue and $14$ red balls. Repeatedly, you take out two balls: if they are the same colour you put back one blue ball, and if they differ you put back one red ball. This continues until a single ball is left. What colour is it? What changes if the bag starts with $13$ red balls instead?

Blue with $14$ red balls; red with $13$. Each step either leaves the number of red balls unchanged or lowers it by two, so its parity never changes. With an even start the red count can reach $0$, so the last ball is blue; with an odd start at least one red always remains, so the last ball is red.