We will show that Ann can win. Ann's strategy is to guarantee that Bob will move upwards a distance of at least $\frac{1}{2^{99}}$ in at most $100$ moves. Here is the strategy: Ann tells Bob $\frac{1}{2^{99}}$. If Bob moves upwards, she repeats the strategy from the start. If Bob moves in any other direction, Ann will say the number $\frac{1}{2^{98}}$ and so on, doubling the value that she previously said, whenever Bob doesnt move upwards. Generally, Ann will say $\frac{1}{2^{i}}$ and if Bob moves upwards, then she starts over from saying $\frac{1}{2^{99}}$, but if Bob moves in any other direction, on the next turn she will say $\frac{1}{2^{i-1}}$. If Bob moved upwards when Ann said $\frac{1}{2^{i}}$, he could've went downwards a maximum of $\frac{1}{2^{99}}+\frac{1}{2^{98}}+\dots+\frac{1}{2^{i-1}}$ so he goes upwards a minimum of $\frac{1}{2^{99}}$. Also, we know that Bob has to go upwards sometime in $100$ moves, so Ann isnt going to say a number greater than $1$ and hence can play this strategy. With this, we have shown that Bob is forced to go upwards a distance of $\frac{1}{2^{99}}$ in maximum $100$ moves so after enough time, he will go upwards a distance of $100$ which clearly means that he will be at a distance greater than $100$ away from the starting point.