The answer is 50. Very important fact in one of strategy for first player is that in the end number of such pairs is always even. For example, players can add or ban one such pairs every theirs move. This works.

Answer: $50$. Strategy for $B$: Label the points as $1,2,...,100$. $B$ pairs $2i-1$ with $2i$. $B$ colours the pair of the point which $A$ coloured before $B'$s move with the same colour so there can be at most $50$ different coloured neighbours. Strategy for $A$: Label the points as $1,2,...,100$. $A$ pairs $2i$ with $2i+1$ for $i=1,2,...,49$ such that each number except $1$ and $100$ has one pair. $A$ colours $1$ as red. After that, if $B$ colours a number, then $A$ colours its pair with the other colour. So there are at least $49$ different coloured pairs. Assume that there are $49$ of them. Then, there mustn't be any different coloured neighbours between two points in distinct pairs. But this is not possible since $49$ is odd. More explicitly, the colours must be $1\rightarrow R, 2\rightarrow R, 3\rightarrow G, 4\rightarrow G, 5\rightarrow R,...,97\rightarrow R,98\rightarrow R, 99\rightarrow G$ hence $100$ is in the same colour with exactly one of $1,99$. This gives that there cannot be $49$ distinct coloured pairs. So $A$ can achieve that there are at least $50$ distinct coloured neighbours.$\blacksquare$