On a plane, there is a complete graph of order $n(\geq 3)$. We color its edges using three colors, and each color is used to color at least one edge. If, regardless of the initial coloring, we can always select $k$ edges in the complete graph, and change the colors of these $k$ edges to a given color among the three, such that the graph can also be connected by the edges of the given color, find the minimum value of $k$.

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Four, Solution: The smallest $k=\left[\frac{n}{3}\right]$.
We first prove $k \geq\left[\frac{n}{3}\right]$.
Assume the three colors are $1, 2, 3$, and the $n$ vertices of the complete graph of order $n$ are divided into three sets $A, B, C$, with $|A|=|B|=\left[\frac{n}{3}\right]$. We color the edges as follows: the edges between points in set $A$ are colored 1, the edges between points in set $B$ are colored 2, the edges between points in set $C$ are colored 3, the edges between points in sets $A$ and $B$ are colored 2, similarly, the edges between points in sets $B$ and $C$ are colored 3, and the edges between points in sets $C$ and $A$ are colored 1. Thus, if the final connected graph formed by the edges colored 1, since the edges from points in set $B$ are all colored 2 or 3, the number of recolored edges $\geq|B|=\left[\frac{n}{3}\right]$; similarly, if the final connected graph formed by the edges colored 2 or 3, then the number of recolored edges $\geq |A|$ (or $|C|) \geq\left[\frac{n}{3}\right]$. Hence, $k \geq\left[\frac{n}{3}\right]$.
Next, we prove $k \leq\left[\frac{n}{3}\right]$. We use induction on $n$.
For $n=3$, the proposition holds.
If the proposition holds for $n-1$ ($n \geq 4$), then for $n$ vertices,
(1) If there is a vertex in the complete graph from which edges of three different colors emanate, by the induction hypothesis, we can obtain a monochromatic connected graph by changing the color of $\left[\frac{n-1}{3}\right]$ edges, which is valid;
(2) If all vertices in the complete graph have edges of at most two different colors emanating from them, let $A$ be the set of vertices from which edges are colored 1 or 2, $B$ be the set of vertices from which edges are colored 2 or 3, and $C$ be the set of vertices from which edges are colored 3 or 1. Without loss of generality, assume $|A| \leq|B| \leq|C|$, then $|A| \leq\left[\frac{n}{3}\right]$. Clearly, $B \neq \varnothing$, so the edges from points in $B$ to points in $C$ are all colored 3, hence $B \cup C$ is connected by color 3. At this point, we can choose any point in $B$, and change the color of the edges from all points in $A$ to this point to 3. Since $|A| \leq\left[\frac{n}{3}\right]$, this is valid.
By induction, the conclusion holds.