8. 113 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 $C_{n}^{j}$, $C_{n}^{i+1}, \cdots, C_{n}^{i+k-1}$ 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 $C_{n}^{j}, C_{n}^{+1}$, $\cdots, C_{n}^{j+k-2}$ form an arithmetic sequence.

Further, find all $n$ that have property $(b)$.
[Proof] Since any two numbers must form an arithmetic sequence, therefore $k \neq 1, k \neq 2$. Now consider three numbers: $C_{n}^{j-1}, C_{n}^{j}, C_{n}^{j+1}(1 \leqslant j \leqslant n-1)$
If they form an arithmetic sequence, then
$$2 C_{n}^{j}=C_{n}^{j-1}+C_{n}^{j+1}$$

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 $n$ that satisfies the condition " $n+2$ is a perfect square", then $n+2=m^{2}$, and since $n, m$ have the same parity, there exists $j$ such that $m=n-2 j$. Thus, (1) holds, and property $(b)$ holds, so when $k=4$, all $n$ with property $(b)$ are
$$n=m^{2}-2 \quad(m \in N, m \geqslant 3)$$

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
$$\begin{array}{l}
C_{2 j+3}^{j}=C_{2 j+3}^{j+3}, \\
C_{2 j+3}^{j}=C_{2 j+3}^{j+1}, \text{ contradiction. }
\end{array}$$

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
$$n=m^{2}-2(m \in N, m \geqslant 3)$$