There are 100 points on a circle that are about to be colored in two colors: red or blue. Find the largest number $k$ such that no matter how I select and color $k$ points, you can always color the remaining $100-k$ points such that you can connect 50 pairs of points of the same color with lines in a way such that no two lines intersect.
1. **Define the problem and notation:**
   - We have 100 points on a circle, labeled \( A_1, A_2, \ldots, A_{100} \).
   - We need to find the largest number \( k \) such that no matter how \( k \) points are colored, the remaining \( 100 - k \) points can be colored in such a way that 50 pairs of points of the same color can be connected with non-intersecting lines.

2. **Initial observation:**
   - If \( k = 50 \), we need to show that no matter how 50 points are colored, we can always color the remaining 50 points to achieve the goal.

3. **Base case:**
   - Consider the case where \( k = 50 \). We need to show that we can always color the remaining 50 points to form 50 non-intersecting pairs of the same color.

4. **Inductive step:**
   - Assume that for \( k = n \), we can always color the remaining \( 100 - n \) points to form 50 non-intersecting pairs of the same color.
   - We need to show that for \( k = n + 1 \), we can still achieve the goal.

5. **Coloring strategy:**
   - If a colored point \( A_j \) has an uncolored neighbor (WLOG \( A_{j+1} \)), color \( A_{j+1} \) the same color as \( A_j \). This forms a pair \( (A_j, A_{j+1}) \) of the same color.
   - Remove \( A_j \) and \( A_{j+1} \) from consideration, reducing the problem to \( k - 1 \) points in \( A_1, \ldots, A_{2k-2} \).

6. **Inductive hypothesis:**
   - By the inductive hypothesis, we can color the remaining \( 100 - (k - 1) \) points to form 50 non-intersecting pairs of the same color.
   - Since we have reduced the problem to \( k - 1 \) points, we can apply the inductive hypothesis to achieve the goal.

7. **Conclusion:**
   - By induction, we have shown that for \( k = 50 \), we can always color the remaining 50 points to form 50 non-intersecting pairs of the same color.

\(\blacksquare\)

The final answer is \( \boxed{ k = 50 } \)