4.45 In a convex polygon, draw all diagonals, and color each side and each diagonal with one of $k$ colors, so that there is no monochromatic closed broken line with the vertices of the polygon as its vertices. How many vertices can such a polygon have at most?
[Solution] Let the number of vertices of the polygon be $n$, then the $n$-sided polygon has $C_{n}^{2}=\frac{1}{2} n(n-1)$ edges and diagonals. If there are $n$ line segments of the same color, then by a theorem in graph theory, there must be a cycle, i.e., a monochromatic closed polyline, which contradicts the given condition. Therefore, the number of line segments of each color does not exceed $n-1$. Thus, we have
$$
\frac{1}{2} n(n-1) \leqslant k(n-1) \text {. }
$$

Solving this, we get $n \leqslant 2 k$.
Next, we show that when $n=2 k$, a coloring method that meets the requirements exists. Take a regular $2 k$-sided polygon $A_{1} A_{2} \cdots A_{2 k}$, and color the polyline $A_{1} A_{2 k} A_{2} A_{2 k-1} \cdots A_{k} A_{k+1}$ with the first color, and color the polyline obtained by rotating this polyline by $\frac{i \pi}{k}$ with the $(i+1)$-th color, $i=1,2, \cdots, k-1$. It is not difficult to see that each line segment is exactly colored with one of the $k$ colors, and there is no monochromatic closed polyline.
In summary, the polygon that meets the conditions has at most $2 k$ sides.