Let $n > k \geqslant 2$ be positive integers. Each of the $n$ points on a circle is colored either red or blue, and it is required that any $k$ consecutive points contain at least one red point. Let the number of valid coloring schemes be denoted as $f_{k}(n)$. Try to find the necessary and sufficient conditions for positive integers $n, k$ such that $f_{k}(n)$ is even.
A coloring scheme is called a good scheme if it satisfies certain conditions. First, consider the number of good schemes for $n$ points on a line, denoted as $g_{k}(n)$.
It is agreed that $g_{k}(0)=1$. When $1 \leqslant n \leqslant k-1$, each point can be freely colored, so $g_{k}(n)=2^{n}$. When $n=k$, only the scheme where all points are blue is not a good scheme, hence $g_{k}(k)=2^{k}-1$.

For $n \geqslant 1$, let the number of good schemes for $n$ points where the last point is red or blue be $g_{k}^{(1)}(n)$ and $g_{k}^{(0)}(n)$, respectively. Since appending a red point to a good scheme $A$ results in a good scheme, $g_{k}^{(1)}(n+1)=g_{k}(n)$. Appending a blue point to a good scheme $A$ results in a good scheme if and only if the last $k-1$ points of $A$ are not all blue. The number of good schemes for $n$ points where the last $k-1$ points are all blue (in which case the $(n-k+1)$-th point from the end must be red) is $g_{k}^{(1)}(n-k+1)=g_{k}(n-k)$. Therefore, $g_{k}^{(0)}(n+1)=g_{k}(k)-g_{k}(n-k)$, and for $n \geqslant k$,
$$
g_{k}(n+1)=g_{k}^{(1)}(n+1)+g_{k}^{(0)}(n+1)=2 g_{k}(n)-g_{k}(n-k) \equiv g_{k}(n-k) \pmod{2},
$$

which means the sequence $g_{k}(n) \pmod{2}$ has a period of $k+1$. Combining the initial values, $g_{k}(n)$ is odd if and only if $n \equiv 0$ or $k \pmod{k+1}$.
A good scheme for $n$ points on a circle is obtained by connecting the ends of a good scheme for $n$ points on a line. A circular scheme is good if and only if the linear scheme is good and the total number of blue points at the beginning and end, $s$, is $\leqslant k-1$. Similarly, for $n \leqslant k-1$, $f_{k}(n)=2^{n}$, and $f_{k}(k)=2^{k}-1$.
For $n \geqslant k+1$, removing the blue points at the beginning and end leaves a good scheme of $n-s$ points with red points at both ends. Since appending a red point to a good scheme $A$ results in a good scheme, the number of such good schemes is $g_{k}(n-s-2)$. Summing over the distributions of the blue points at the beginning and end, we get
$$
f_{k}(n)=\sum_{s=0}^{k-1} \sum_{a=0}^{s} g_{k}(n-s-2)=\sum_{s=0}^{k-1}(s+1) g_{k}(n-s-2) \equiv \sum_{i=0}^{\left[\frac{k-1}{2}\right]} g_{k}(n-2 i-2) \pmod{2}.
$$
$g_{k}(n-2 i-2)$ is odd if and only if $n-2 i-2 \equiv 0$ or $-1 \pmod{k+1}$, i.e., $n \equiv 2 i+1$ or $2 i+2 \pmod{k+1}$. As $i$ ranges from $0,1, \cdots, \left[\frac{k-1}{2}\right]$, $\{2 i+1,2 i+2\}$ covers $S=\left\{1,2, \cdots, 2\left[\frac{k+1}{2}\right]\right\}$. When $k$ is odd, $S$ is a complete residue system modulo $k+1$, so there is exactly one $i$ such that $n \equiv 2 i+1$ or $2 i+2 \pmod{k+1}$, meaning the sum of $\left[\frac{k+1}{2}\right]$ terms in the above $f_{k}(n)$ has exactly one odd term, hence $f_{k}(n)$ is always odd for $n \geqslant k$. When $k$ is even, $S$ is a reduced residue system modulo $k+1$ excluding 0, so $f_{k}(n)$ is always odd for $n \geqslant k$ and $n$ not divisible by $k+1$, and $f_{k}(n)$ is even when $n$ is divisible by $k+1$.
Therefore, $f_{k}(n)$ is even if and only if $k$ is even and $n$ is divisible by $k+1$.