Example 21([31.2]) Let $n \geqslant 3$ be a positive integer, and let $E$ be a set of $2 n-1$ distinct points on the same circle. Some of the points in $E$ are colored black, and the rest are not colored. If there is at least one pair of black points such that one of the two arcs with these points as endpoints has exactly $n$ points of $E$ in its interior (not including the endpoints), then this coloring is called good. Find the smallest $k$ such that any coloring of $k$ points in $E$ is good.
Let the $2n-1$ points be numbered clockwise as $1, 2, 3, \cdots, 2n-1$.
(i). Take the modulus $m = 2n-1$, and under modular arithmetic, assign each point infinitely many labels. Label the $2n-1$ points cyclically starting from point 1 in the clockwise direction: 1, 2, 3, ..., $2n-2, 2n-1; 2n, 2n+1, \cdots$. Labels in the reverse direction can also be considered, i.e., label the points counterclockwise from point 1 as: $1, 0, -1, -2, -3, \cdots, -(2n-3); -(2n-2)$, $-(2n-1), -2n, -(2n+1), \cdots$ (but this is not used here). Thus, each point has infinitely many labels, but if $i \equiv j \pmod{2n-1}$, $i$ and $j$ represent the same point. That is, numbers in the same congruence class modulo $2n-1$ represent the same point. This way, the discussion of the problem can utilize modular arithmetic.
(ii). Fix a point $i$ and color it black. In the clockwise direction, the first point that, when colored black, forms a good coloring with point $i$ is $i + (n-2)$. Using modular arithmetic, reorder these points so that in the new order, any coloring with adjacent black points is good. For a fixed $i$, consider the points
$$a_{i, r} = i + r(n-2), \quad r = 0, 1, \cdots$$

Thus, a coloring method is good if there is some $r$ such that the adjacent points $a_{i, r}$ and $a_{i, r+1}$ in the new order are both black. Therefore, if coloring any $k$ points in $E$ results in at least two adjacent points being black in this new order, then the coloring method is good. The problem then becomes finding the smallest such $k$. However, the first issue to address is whether, for a fixed $i$, such as $i = 1$, the points $a_{1, r} = 1 + r(n-2)$ $(r = 0, 1, \cdots)$ can represent all $2n-1$ different points. If not, how should we proceed further?
(iii). Now fix $i = 1$, and the points $a_{1, r}$ and $a_{1, s}$ represent the same point if and only if
$$a_{1, r} \equiv a_{1, s} \pmod{2n-1}, \quad (s-r)(n-2) \equiv 0 \pmod{2n-1}$$

Note that
$$(2n-1, n-2) = (3, n-2) = (2n-1, 3)$$

We need to consider two cases:
(a) When $(2n-1, n-2) = (3, 2n-1) = 1$, i.e., $n \neq 3l + 2$, sequentially connect the points $a_{1, r} = 1 + r(n-2)$ and $a_{1, r+1} = 1 + (r+1)(n-2)$. If $r = 0, 1, \cdots, 2n-2$, this forms a closed broken line connecting all $2n-1$ points. In this case, coloring any $n$ points among $a_{1, r} (0 \leq r \leq 2n-2)$ will ensure that there are two adjacent points $a_j$ and $a_{j+1}$ that are both black, so the coloring method is good. If the points $a_{1,1}, a_{1,3}, \cdots, a_{1,2n-3}$ (a total of $n-1$ points) are colored black, the coloring method is not good. Therefore, $k = n$.
(b) When $(2n-1, n-2) = (3, 2n-1) = 3$, i.e., $n = 3l + 2$, the condition for $a_{1, r}$ and $a_{1, s}$ to represent the same point
$$(s-r)(n-2) \equiv 0 \pmod{2n-1}$$

becomes (why?)
$$(s-r) \equiv 0 \pmod{(2n-1)/3}$$

We will find that, for a fixed $i$, connecting $a_{i, r}$ and $a_{i, r+1}$ as $r$ goes to $r+1$ does not connect all $2n-1$ points but only $(2n-1)/3 = 2l+1$ points into a closed curve. It is easy to see that all points will belong to three such closed broken lines, and each point belongs to only one. These three closed broken lines can be taken as
$$a_{i, r}, \quad r = 0, 1, \cdots, (2n-4)/3, i = 1, 2, 3$$

Thus, when any $n-1 = 3l+1$ points among the $2n-1$ points are colored black, there must be $l+1$ black points on one of these closed broken lines (say $i = 1$), so there must be two adjacent points $a_{1, r}$ and $a_{1, r+1}$ that are both black, making the coloring method good. However, coloring the points $a_{i, r} (r = 1, 3, 5, \cdots, (2n-7)/3 = 2l-1; i = 1, 2, 3)$ (a total of $n-2 = 3l$ points) is not good. Therefore, $k = n-1$.