Find a positive integer $k$ such that

(a) For any positive integer $n$, there does not exist $j$ satisfying $0 \leqslant j \leqslant n - k + 1$, and $C_{n}^{j}, C_{n}^{j+1}, \cdots, C_{n}^{j+k-1}$ form an arithmetic sequence;

(b) There exists a positive integer $n$ such that there is $j$ satisfying $0 \leqslant j \leqslant n - k + 2$, and $C_{n}^{j}, C_{n}^{j+1}, \cdots, C_{n}^{j+k-2}$ form an arithmetic sequence. Further, find all $n$ with property (b).
Proof: Since any two numbers must form an arithmetic sequence, let $k=7=$ $1, k \neq 2$. Now consider three numbers:
$$
C_{n}^{j-1}, C_{n}^{j}, C_{0}^{j+i}(1 \leqslant j \leqslant n-1) .
$$

If they form an arithmetic sequence, then
$$
2 C_{n}^{j}=C_{n}^{j-1}+C_{n}^{j+1} \text {. }
$$

This implies $n+2=(n-2 j)^{2}$,
i.e., $n+2$ is a perfect square.
This proves that when $k=3$, $(a)$ does not hold, hence $k \neq 3$. It also proves that when $k=4$, $(b)$ holds.

Conversely, if there exists a positive integer $n$ satisfying the condition “$n+2$ is a perfect square”, then $n+2=m^{2}$. Since $n$ and $m$ have the same parity, there exists $j$ such that $m=n-2 j$. Thus, (1) holds, and property $(b)$ holds. Therefore, when $k=4$, all $n$ with property $(b)$ are
$$
n=m^{2}-2(m \in \mathbf{N}, m \geqslant 3) \text {. }
$$

Next, we prove that when $k=4$, property $(a)$ also holds.
If $C_{n}^{j}, C_{n}^{j+1}, C_{n}^{j+2}, C_{n}^{j+3}$ form an arithmetic sequence, then by (1) we have
$$
\begin{array}{l}
n=(n-2(j+1))^{2}-2=(n-2(j+2))^{2}-2, \\
|n-2 j-2|=|n-2 j-4|, \\
n-2 j-2=-(n-2 j-4), \\
n=2 j+3 .
\end{array}
$$

The original sequence is $C_{2 j+3}^{j}, C_{2 j+3}^{j+1}, C_{2 j+3}^{j+2}, C_{2 j+3}^{j+3}$.
But $C_{2 j+3}^{j}=C_{2 j+3}^{j+3}$,
so $C_{2 j+3}^{j}=C_{2 j+3}^{j+1}$, which is a contradiction.
Therefore, when $k=4$, $(a)$ holds, and when $k=5$, $(b)$ does not hold, i.e., $k<5$.

In summary, the required positive integer $k=4$, and all $n$ with property $(b)$ are
$$
n=m^{2}-2(m \in \mathbf{N}, m \geqslant 3) .
$$