To each situation we associate the number $M$, the maximal amount of balls in a pile. With each move, $M$ can increase by at most 1. If the first player (A) can ensure that $M$ always increases, then he wins by parity since $B$ leaves $M=20$ and loses. $A$ can do this by making sure that he always leaves a situation with at least two maximal piles. Let's see why. - If $B$ causes $M\rightarrow M+1$ in at least two places, then $A$ can increase these piles to $M+2$ and some other ones. - Suppose $B$ effectuates $M\rightarrow M+1$ in only one pile, say pile $k$. Before his move, there was a different pile with $M$ balls, say in position $l$. Now $A$ can increase all piles except pile $k$. Then the new maximum $M+1$ appears in at least two piles, namely the ones numbered $k,l$. It follows that $A$ has a winning strategy. PS. This works when 21 is any odd number. For even numbers B has a winning strategy, namely the same one as described above.