9.37 Let $n \geqslant 3$ be a natural number. 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 between them contains exactly $n$ points of $E$ in its interior, then this coloring is called "good". If every coloring of $k$ points in $E$ is good, find the minimum value of $k$.

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[Solution 1] In a counterclockwise direction, label the given $2n-1$ points as $A_{1}, A_{2}, \cdots, A_{2n-1}$, and agree that when $k > 2n-1$, $A_{k} = A_{k-2n+1}$. For any point $A_{i}$, there are exactly two points that are $n$ points away from it, namely $A_{i+n+1}$ and $A_{i+n-2}$. We call these two points the "associated points" of $A_{i}$. Clearly, if there is a pair of black points that are associated points, then this coloring method is good.

First, we prove that the minimum value of $k$ is no more than $n$. If not, then there exists a coloring method that colors $n$ points in $E$ and is not good. At this time, the two associated points of each black point are all uncolored, so there are $2n$ uncolored points (including repeated counts). However, each uncolored point is at most the associated point of two black points, so there are at least $n$ different uncolored points, thus the total number of points is at least $2n$, which is impossible.

Second, if we color the first $n-2$ points and leave the rest uncolored, then any two black points are not associated points. Therefore, the minimum value of $k$ is no less than $n-1$.

Next, we examine the cases when $n=3, 4, 5$ to gain information about the minimum value of $k$ in the general case.

When $n=3$, $2n-1=5$. Clearly, two points are associated points if and only if they are adjacent. Therefore, when two non-adjacent points are colored black, the coloring method is not good. Thus, the minimum value of $k$ is $3=n$.

When $n=4$, $2n-1=7$. Two points are associated points if and only if there is one point between them. Therefore, when $A_{1}, A_{4}, A_{5}$ are colored black, any two black points are not associated points. Thus, the minimum value of $k$ is $4=n$.

When $n=5$, $2n-1=9$. Divide the 9 known points into 3 groups:
$\left\{A_{1}, A_{4}, A_{7}\right\},\left\{A_{2}, A_{5}, A_{8}\right\},\left\{A_{3}, A_{6}, A_{9}\right\}$.

It is not difficult to see that the two associated points of any point are exactly the other two points in the same group. If 4 points out of the 9 are colored, then by the pigeonhole principle, there must be two black points in the same group, thus they are associated points. Thus, the minimum value of $k$ is $4=n-1$.

The above discussion shows that the minimum value of $k$ is sometimes $n$ and sometimes $n-1$, depending on whether $2n-1$ is divisible by 3. We will prove this separately.
(1) When $3 \mid (2n-1)$, let $2n-1=3m$ and divide the given points into 3 groups:
$$
S_{i}=\left\{A_{j} \mid j=3k+i, k=0,1, \cdots, m-1\right\}, i=1,2,3.
$$

Note that $m$ is odd, so
$$
n-2=3 \cdot \frac{m-1}{2}=3\left[\frac{m}{2}\right], n+1=3\left(\left[\frac{m}{2}\right]+1\right)
$$

are both multiples of 3, thus $A_{i+n+1}$ and $A_{i+n-2}$ are in the same group as $A_{i}$. If $n-1=3\left[\frac{m}{2}\right]+1$ points are colored black, then by the pigeonhole principle, there must be at least $\left[\frac{m}{2}\right]+1$ black points in one of the 3 groups. Thus, like the argument in the first paragraph, we can prove that there must be two black points in this group that are associated points. Therefore, any coloring method that colors $n-1$ points is good. So, the minimum value of $k$ is $n-1$.
(2) When $3 \nmid (2n-1)$, let $2n-1=3m \pm 1$. It is easy to see that $m$ is even. Let
$$
S=\left\{A_{j} \mid j=3k+1, k=0,1,2, \cdots, n-2\right\},
$$

then the associated points of any point in $S$ are not in $S$. In fact, from $2n-1=3m \pm 1$ we can deduce
$$
\left\{\begin{array}{l}
n-2=\frac{3m}{2}-1, \\
n+1=\frac{3m}{2}+2,
\end{array} \quad \left\{\begin{array}{l}
n-2=\frac{3m}{2}-2, \\
n+1=\frac{3m}{2}+1,
\end{array}\right.\right.
$$
4 numbers are not multiples of 3. So, if $A_{i} \in S$, then its two associated points $A_{i+n+1}$ and $A_{i+n-2}$ are not in $S$. Now, if we color all $n-1$ points in $S$ black, then this coloring method is not good. So, the minimum value of $k$ in this case is $n$.
In summary, we get
$$
k_{\text{min}}=\left\{\begin{array}{l}
n-1, \text{ when } 3 \mid (2n-1), \\
n, \quad \text{ when } 3 \nmid (2n-1).
\end{array}\right.
$$
[Solution 2] Like in Solution 1, we label the $2n-1$ given points as $A_{1}, A_{2}, \cdots, A_{2n-1}$ and introduce the concept of associated points. Now, connect a line between each pair of associated points, then we get a graph $G$, where the degree of each point is 2. Therefore, the even graph $G$ is a cycle or a union of several disjoint cycles. By symmetry, each cycle has the same number of vertices. So, the number of disjoint cycles must be odd. Since any two associated points are separated by two points, we know that $A_{i}$ and $A_{i+3}$ are in the same cycle, so the number of cycles is at most 3. If there are exactly 3 cycles, then since each cycle has the same number of vertices, we have $3 \mid (2n-1)$. Conversely, if $3 \nmid (2n-1)$, then $3 \mid (n-2)$, $3 \mid (n+1)$. Therefore, the subgraph with any of the following 3 sets of vertices
$\{1,4, \cdots, 2n-3\},\{2,5, \cdots, 2n-2\},\{3,6, \cdots, 2n-1\}$ is a cycle. Therefore, the graph $G$ can be decomposed into 3 cycles if and only if $3 \mid (2n-1)$. When $3 \nmid (2n-1)$, the graph $G$ is a cycle.

Clearly, a coloring method is good if and only if there is a pair of adjacent vertices in graph $G$ that are both black. When $3 \nmid (2n-1)$, the graph $G$ is a cycle, and at most $n-1$ vertices can be selected such that no two are adjacent. So, if we color such a group of $n-1$ points black, then the coloring method is not good; if we color any $n$ points black, there must be two adjacent vertices, so the coloring method is always good. Therefore, the minimum value of $k$ is $n$. When $3 \mid (2n-1)$, the graph $G$ has 3 cycles, each with $m=\frac{1}{3}(2n-1)$ vertices. In any cycle, at most $\left[\frac{m}{2}\right]$ vertices can be selected such that no two are adjacent. So, if $\left[\frac{m}{2}\right]+1$ vertices in a cycle are colored black, there must be two adjacent vertices. Therefore, when $n-1=3\left[\frac{m}{2}\right]+1$ points are colored black, by the pigeonhole principle, at least one of the 3 cycles has at least $\left[\frac{m}{2}\right]+1$ black points, thus such a coloring method is always good. If the first $n-2$ known points are colored black, then any two black points are not adjacent vertices in the graph, so this coloring method is not good. Therefore, the minimum value of $k$ in this case is $n-1$.
In summary, we get, $k_{\text{min}}=n$, when $3 \nmid (2n-1)$; $k_{\text{min}}=n-1$, when
$3 \mid (2n-1)$.