C39 (33-3, China) Given 9 points in space, where no 4 points are coplanar, a line segment is drawn between every pair of points. These line segments can be colored red or blue, or left uncolored. Find the smallest value of $n$ such that if any $n$ of these line segments are each arbitrarily colored red or blue, the set of these $n$ line segments will necessarily contain a triangle with all sides of the same color.
Solution: Suppose the number of colored segments is at least 33, then the number of uncolored segments is at most
$$
\mathrm{C}_{9}^{2}-33=36-33=3 \text { (segments). }
$$

If point $A_{1}$ leads to uncolored segments, remove $A_{1}$ and the segments it leads to. If there is still a point $A_{2}$ leading to uncolored segments in the remaining graph, remove $A_{2}$ and the segments it leads to, and continue this process. Since the number of uncolored segments is at most 3, at most 3 vertices (and the segments they lead to) can be removed, leaving at least 6 points, with each pair of points connected by a red or blue line. It is easy to see that there must be a monochromatic triangle among them.

The following example shows that if the number of colored segments is less than 33, there may not be a monochromatic triangle. As shown in the figure, the edges between $v_{1}$ and $v_{9}$, $v_{2}$ and $v_{8}$, $v_{3}$ and $v_{7}$, $v_{4}$ and $v_{6}$ are not colored, and the solid lines in the graph represent red, while the dashed lines represent blue. The dashed lines between $v_{1}, v_{9}$ and $v_{2}, v_{8}$ indicate that $v_{1} v_{2}, v_{1} v_{8}, v_{9} v_{2}, v_{9} v_{8}$ are all dashed lines, and similarly for the rest. Then there is no monochromatic triangle in the graph.