Solution:

The answer is $\mathbf{( A )}$. Suppose we start from an odd number $n$: then we can write it as $n=2k+1$, and multiplying by 51 and adding 50 we find $102k+101=100k+100+(2k+1)$. Now $100k+100$ is a number ending in two zeros, so the last two digits of $51n+50$ are the same as those of $n$, and the game stays fixed forever on $n$.

If instead we start from an even number $n=2k$, the rule gives $102k+50$, which ends with the same two digits as $2k+50$: so if $2k$ is less than 50 we simply find $2k+50$, and otherwise we find $2k-50$ because of the carry. At the next step, the same reasoning tells us that $2k+50$ will become $2k+50-50=2k$ (since $2k+50$ is certainly at least equal to 50), and similarly $2k-50$ will become $2k-50+50=2k$: the game repeats every two turns, and therefore the different numbers we can obtain in the course of a game are at most 2.