C35 (31-2, Czechoslovakia) Let $n \geqslant 3$, consider a set $E$ of $2 n-1$ distinct points on a circle. If some points in $E$ are colored black, such that there are at least two black points, and one of the two arcs with these black points as endpoints contains exactly $n$ points of $E$ in its interior, then this coloring method is called "good". If every coloring method of $k$ points in $E$ being colored black is good, find the minimum value of $k$.


Let the points in $E$ be denoted as $1, 2, \cdots, 2n-1$, and connect point $i$ with $i+(n-1)$ (with the convention that $j+(2n-1)k, k \in \mathbb{Z}$, represents the same point as $j$). This results in a graph $G$, where each vertex in $G$ has a degree of 2 (connected to two points), and points with a difference of 3 are adjacent to the same point.
Since each vertex in graph $G$ has a degree of 2, $G$ consists of one or several cycles.
When $3 \nmid 2n-1$, each point $j$ in $1, 2, \cdots, 2n-1$ can be expressed as $3k$ (i.e., the equation $3x \equiv j \pmod{2n-1}$ has a solution), so graph $G$ is a cycle of length $2n-1$. On this cycle, $n-1$ points can be selected such that no two points are adjacent, and at most $n-1$ such points can be selected.

When $3 \mid 2n-1$, graph $G$ consists of three cycles of length $\frac{2n-1}{3}$, with the vertex sets of the cycles being
$$
\begin{array}{l}
\left\{3k+1, k=0,1, \cdots, \frac{2n-4}{3}\right\}, \\
\left\{3k+2, k=0,1, \cdots, \frac{2n-4}{3}\right\}, \\
\left\{3k, k=1,2, \cdots, \frac{2n-1}{3}\right\}.
\end{array}
$$

On each cycle, at most $\frac{1}{2}\left(\frac{2n-1}{3}-1\right)=\frac{n-2}{3}$ points can be selected such that no two points are adjacent, totaling $n-2$ points that are not adjacent.
In summary,
$$
k_{\text{min}}=\left\{\begin{array}{l}
n, \text{ when } 3 \nmid 2n-1, \\
n-1, \text{ when } 3 \mid 2n-1.
\end{array}\right.
$$