Suppose at step n you add the following numbers, 2n-1, 2(2n-1), 4(2n-1)… (2^k)(2n-1) for all k. Then you take away the number n.
This procedure adds infinitely many at every step, and only takes one away, so by your argument it should only ever grow and end up as infinity. However, its pretty clear that every number gets added once and then taken away once, so we end up with nothing.
Technically speaking its a slight modification, since usually the paradox involves adding a finite number than taking away 1, rather than infinitely many.
Minor note: this is the Ross-Littlewood paradox.
Technically speaking its a slight modification, since usually the paradox involves adding a finite number than taking away 1, rather than infinitely many.