8. Given positive integers $n, k$, where $n$ is even, $k \geqslant 2$, and $n > 4k$. There are $n$ points on a circle. If $\frac{n}{2}$ non-intersecting chords inside the circle have endpoints exactly at these $n$ points, then these chords are called a "matching". Find the largest integer $m$ such that for any matching, there exist $k$ consecutive points on the circle, such that at least $m$ chords in this matching have both endpoints among these $k$ points.
（Zhai Zhenhua, problem contributor）
8. $\left\lceil\frac{k}{3}\right\rceil$ ( $\lceil x\rceil$ denotes the smallest integer not less than the real number $x$). Let $m=\left\lceil\frac{k}{3}\right\rceil$.
It is easy to see that $k \geqslant 2 m$, and $3 m-2 \leqslant k \leqslant 3 m$.
First, we provide an example of a matching where any $k$ consecutive points contain at most $m$ chords.

Assume these $n$ points are the vertices of a regular $n$-gon. Since $n$ is even and $n>4 k \geqslant 8 m$, we can divide these $n$ points into six consecutive segments $M_{1}, M_{2}, X_{1}, M_{3}, M_{4}, X_{2}$, with the number of points being $2 m, 2 m, x, 2 m, 2 m, x$ respectively. In each $M_{i}$, we can connect $m$ mutually parallel chords by pairing the first and last points. Between the points $X_{1}$ and $X_{2}$, we can pair and connect $x$ mutually parallel chords, which are called "long". All the chords do not intersect inside the circle, forming a matching. In any $k$ consecutive points $K$, it is impossible to include both endpoints of a long chord, otherwise, it would include at least $4 m+2 > k$ points. If $K$ includes a chord in $M_{i}$, then $K$ must include the middle two points of $M_{i}$. Thus, $K$ cannot include chords from three $M_{i}$ segments, otherwise, by symmetry, assume $K$ includes chords from $M_{1}, M_{2}, M_{3}$, then $K$ includes the middle two points of $M_{1}$ and $M_{3}$. Therefore, $K$ includes at least half of the points, $|K| \geqslant \frac{n}{2} > k$, which contradicts the condition. Hence, $K$ contains at most two chords from $M_{i}$. If $K$ contains a chord from one $M_{i}$, then $K$ contains no more than $m$ chords. If $K$ contains chords from two $M_{i}$ segments, assume $K$ contains chords from $M_{i}$ and $M_{i+1}$, $\left|K \cap M_{i}\right|=s,\left|K \cap M_{i+1}\right|=t$, then $s+t \leqslant k$, and by the continuity of $K$, $K \cap M_{i}$ is the last $s$ points, and $K \cap M_{i+1}$ is the first $t$ points, and $s>m, t>m$, so $K$ contains exactly $(s-m)+(t-m)$ chords. At this time,
$$
\begin{array}{l}
(s-m)+(t-m)=s+t-2 m \\
\leqslant k-2 m \leqslant 3 m-2 m=m .
\end{array}
$$

Next, we prove that for any matching, there are always $k$ consecutive points containing at least $m$ chords.

For a chord, define its length as the number of points (including the endpoints) on the shorter arc (or semicircle) it divides. Label these $n$ points in a clockwise direction starting from a certain point as $P_{1}, P_{2}, \cdots, P_{n}$.
We discuss in three cases.
(1) There exists a chord with length $l \in [2 m, k]$.
We can take $k$ consecutive points, including the two endpoints of this chord and all $l$ points on one side. Since all chords do not intersect inside the circle, the points on one side of this chord can only pair with points on the same side. Therefore, these $k$ consecutive points contain at least $\frac{l}{2} \geqslant m$ chords.
(2) No chord has a length in the interval $[2 m, k]$, but there exists a chord with a length greater than $k$.

Take the chord with the smallest length among those with a length greater than $k$, assume it is $P_{1} P_{l}(l>k)$. Take $k$ consecutive points $P_{2}, P_{3}, \cdots, P_{k+1}$, in these $k$ points, there are at least $m$ chords.

In fact, divide $P_{2}, P_{3}, \cdots, P_{l-1}$ into several segments as follows: assume $P_{2}$ is connected to $P_{i}$, then the points between $P_{2}$ and $P_{i}$ are called the first segment, then assume $P_{i+1}$ is connected to $P_{j}$, then $i+1<j$, the points between $P_{i+1}$ and $P_{j}$ are called the second segment, and so on. Assume a total of $s$ segments, with the number of points in each segment being $l_{1}, l_{2}, \cdots, l_{s}$.
By $l_{i}<l$ and the assumption of $l$, we have $l_{i} \leqslant 2 m-2$.
Thus, $\sum_{i=1}^{s} l_{i}=l-2 \geqslant k-1 \geqslant 2 m-1$.
Therefore, there exists a unique positive integer $i_{0}\left(1 \leqslant i_{0}<s\right)$, satisfying $\sum_{j=1}^{i_{0}} l_{j} \leqslant 2 m-2<\sum_{j=1}^{i_{0}+1} l_{j}$.
Assume $\sum_{j=1}^{i_{0}} l_{j}=2 m-2 t, l_{i_{0}+1}=2 p$. Then
$t<p \leqslant m-1$.
By $k+(2 m-2 t) \geqslant p+t$
$\Leftrightarrow k+t \geqslant p+2 m \Leftarrow k+1 \geqslant 3 m-1$, we know that $P_{2}, P_{3}, \cdots, P_{k+1}$ includes the points in the first $i_{0}$ segments and at least $p+t$ points in the $(i_{0}+1)$-th segment. In the first $i_{0}$ segments, there are exactly $m-t$ arcs. In $p$ pairs of points, taking $p+t$ points at least includes $t$ pairs, so in the $(i_{0}+1)$-th segment, at least $t$ chords are included. Therefore, $P_{2}, P_{3}, \cdots, P_{k+1}$ contains at least $m$ arcs.
(3) All chords have a length less than $2 m$.
Take the longest chord, assume it is $P_{1} P_{l_{1}}\left(l_{1}<2 m\right)$. First, consider the chord $P_{l_{1}+1} P_{j}$.

By the maximality of $l_{1}$, $P_{1} P_{l_{1}}$ cannot be contained on the shorter arc side of $P_{l_{1}+1} P_{j}$, so
$$
j=l_{1}+l_{2}, l_{2}<2 m .
$$

Then consider the chord connected to $P_{j+1}$, and so on, we can divide all $n$ points into several segments, each segment has fewer than $2 m$ points, and the first and last points of each segment are connected by a chord.

Similar to the proof in (2), we know that taking $k$ consecutive points starting from $P_{1}$ must contain at least $m$ chords.