4. Given 100 different points on a circle, determine the largest positive integer $k$: such that any $k$ of these 100 points can be arbitrarily colored red or blue, and the remaining points can be appropriately colored red or blue, so that 50 segments can be drawn using these 100 points as endpoints, with no two segments sharing a common point, and the endpoints of each segment being the same color.
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Replace 100 points with $2 m$ points, and label them in a clockwise direction as $A_{1}, A_{2}, \cdots, A_{2 m}$. Each point is colored either red or blue. If it is possible to pair these $2 m$ points with $m$ line segments such that the conditions stated in the problem are met, then the coloring method is considered good. Let $S=\left\{A_{1}, A_{3}, \cdots, A_{2 m-1}\right\}$ and $T=\left\{A_{2}, A_{4}, \cdots, A_{2 m}\right\}$. 

Lemma: A coloring method of $2 m$ points in red and blue is good if and only if the number of red points in $S$ equals the number of red points in $T$.

Necessity. Suppose there exists a pairing of the points that meets the requirements, then the red points in $S$ must be paired with the red points in $T$ (if a line segment $l$ has both endpoints in $S$, then each arc on either side of $l$ contains an odd number of points, so there must be a line segment intersecting with line segment $l$). Similarly, the red points in $T$ must be paired with the red points in $S$, thus the required pairing of line segments provides a one-to-one correspondence between the red points in $S$ and the red points in $T$.

Sufficiency. We prove by induction on $m$ that when the number of red points in $S$ equals the number of red points in $T$, there exists a pairing of line segments that meets the requirements. Note that when $m=1$, by the assumption, the two points are of the same color, and a line segment can be drawn between them, so the conclusion is obviously true.

Assume $m \geqslant 2$, and the conclusion holds for $m-1$. Consider the case for $m$. In this case, there exist two adjacent points of the same color, otherwise, the points would be alternately red and blue, which would mean all points in $S$ are of one color and all points in $T$ are of the other color, contradicting the assumption. Without loss of generality, assume $A_{2 m-1}$ and $A_{2 m}$ are of the same color, and connect the line segment $A_{2 m-1} A_{2 m}$. The remaining $2 m-2$ points satisfy the conditions of the induction hypothesis, and any line segment connecting two of these points will not intersect with the line segment $A_{2 m-1} A_{2 m}$. By the induction hypothesis, for these $2 m-2$ points, $m-1$ line segments can be drawn to pair them and meet the requirements. Adding the line segment $A_{2 m-1} A_{2 m}$ gives a pairing of line segments for the $2 m$ points that meets the requirements. The lemma is proved.

Returning to the original problem. If $k \geqslant 51$, we can color all 50 points in $S$ red and $k-50$ points in $T$ blue. No matter how the remaining points are colored, the necessary and sufficient condition of the lemma cannot be met. Therefore, $k \geqslant 51$ does not satisfy the requirements.

If $k=50$, assume any 50 points are colored, with $a$ red points in $S$ and $b$ red points in $T$, and without loss of generality, assume $a \geqslant b$. Then there are at least $a-b$ points in $T$ that are not colored (otherwise, the number of uncolored points in $T$ $\leqslant a-b-1$, then the number of colored points $\geqslant a+50-(a-b-1)=51+b \geqslant 51$, which is a contradiction). Thus, we can color the $a-b$ uncolored points in $T$ red, and the rest of the uncolored points blue. This way, $S$ and $T$ each have $a$ red points, meeting the condition of the lemma, so the coloring method is good.

In summary, the maximum value of $k$ is 50.

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