$(119-2)/9=13$ moves. So $A$ has $7$ moves and $B$ has $6$ moves. That means $B$ erases $6\cdot9=54$ numbers. Some strategies for $B$ to lower $A$'s score is to erase numbers from extremes, either the first $54$ numbers, the last $54$ numbers, or the first $27$ and the last $27$ numbers. In these three examples $k=64$ and they are counterexamples for $k=65$. We will show that $64$ is the answer. Let's take the pairs $(x, x+64)$. Then we have the pairs $(1, 65), (2, 66),..., (55, 119)$. Notice that $9$ numbers are not in these pairs: $56, 57,..., 63, 64$. The first move of $A$ will be erasing the aforementioned $9$ numbers. After this $A$ and $B$ have the same number of moves. Since the next move belongs to $B$, because he will erase $9$ numbers, he will be the first to erase only a number from a $(x, x+64)$ pair(s), $A$ can erase the other number from the pair(s) in the following move. It doesn't matter if $B$ erases a $(x, x+64)$ pair. Other numbers erased by $A$ will also be such pairs. Then the remaining two numbers will be a $(x, x+64)$ pair. This means that $A$ can be sure that the difference between the remaining pair is not smaller than $64$.