It's very simple actually. Instead of integers, suppose we have a bag which contains two white balls and two black balls. You draw one, but don't replace it. So there is 3 balls left, two of which will be the opposite colour to the one you just drew. The second draw will always be biased towards the opposite colour. In your case you are enumerating all possible draws, so obviously you should get more "opposites" than "sames".
Oh, and as n increases it converges to 50/50 as you said. Why? Because the effect of removing one ball is proportionally smaller as n increases.
