1. Labeling the Points:
We label the points on the circle as $1, 2, \ldots, 100$.

2. Borya's Strategy:
- Borya pairs the points as $(2i-1, 2i)$ for $i = 1, 2, \ldots, 50$.
- If Asya colors a point $2i-1$ or $2i$, Borya colors the other point in the pair with the same color.
- This strategy ensures that each pair $(2i-1, 2i)$ is monochromatic, resulting in no adjacent pairs of distinct colors within each pair.
- Since there are 50 such pairs, Borya can ensure that there are at most 50 pairs of adjacent points with distinct colors.

3. Asya's Strategy:
- Asya pairs the points as $(2i, 2i+1)$ for $i = 1, 2, \ldots, 49$, and considers the pairs $(1, 2)$ and $(99, 100)$ separately.
- Asya colors point 1 with color $R$ (red).
- If Borya colors a point, Asya colors its pair with the opposite color.
- This strategy ensures that each pair $(2i, 2i+1)$ has distinct colors, resulting in 49 pairs of adjacent points with distinct colors.
- Additionally, since the circle is closed, the points 1 and 100 are also adjacent. If Borya colors point 100, Asya can color it with the opposite color of point 1, ensuring one more pair of distinct colors.
- Therefore, Asya can ensure at least 50 pairs of adjacent points with distinct colors.

4. Conclusion:
- Borya can ensure that there are at most 50 pairs of adjacent points with distinct colors.
- Asya can ensure that there are at least 50 pairs of adjacent points with distinct colors.
- Therefore, the maximal number of such pairs Asya can guarantee to obtain, no matter how Borya plays, is 50.

The final answer is $\boxed{50}$