Four. (50 points) Given a prime number $p$ and a positive integer $k$ $(k<p)$. Find all positive integers $n(n \geqslant p-1)$, such that
$$
\mathrm{C}_{n}^{i}, \mathrm{C}_{n}^{i+1}, \cdots, \mathrm{C}_{n}^{i+p-1}(i=1,2, \cdots, n-p+1)
$$

contain exactly $k$ numbers divisible by $p$.
Four, first prove a lemma.
Lemma For any positive integers $m, n (m < n)$, let $n = (a_1 a_2 \cdots a_t)_p, m = (b_1 b_2 \cdots b_t)_p$ (allowing $b_1 = b_2 = \cdots = b_i = 0$). Then $p \nmid \binom{n}{m}$ if and only if for all $i = 1, 2, \cdots, t$, we have $b_i \leq a_i$.

Proof Use $v_p\left(\binom{n}{m}\right)$ to denote the power of the prime $p$ in $\binom{n}{m}$, and $[x]$ to denote the greatest integer not exceeding the real number $x$. Then
$$
\begin{array}{l}
v_p\left(\binom{n}{m}\right) = v_p\left(\frac{n!}{m!(n-m)!}\right) \\
= \sum_{j=1}^{+\infty} \left( \left[\frac{n}{p^j}\right] - \left[\frac{m}{p^j}\right] - \left[\frac{n-m}{p^j}\right] \right) \\
= \sum_{j=1}^{t-1} \left( \left[\frac{n}{p^j}\right] - \left[\frac{m}{p^j}\right] - \left[\frac{n-m}{p^j}\right] \right).
\end{array}
$$
(1) Sufficiency.

If for all $i = 1, 2, \cdots, t$, we have $b_i \leq a_i$, then for any $j (j = 1, 2, \cdots, t-1)$, we have
$$
\begin{array}{l}
\left[\frac{n}{p^j}\right] - \left[\frac{m}{p^j}\right] - \left[\frac{n-m}{p^j}\right] \\
= (a_1 a_2 \cdots a_{t-j})_p - (b_1 b_2 \cdots b_{t-j})_p - \\
\quad \left( (a_1 - b_1)(a_2 - b_2) \cdots (a_{t-j} - b_{t-j}) \right)_p \\
= 0.
\end{array}
$$
Thus, $v_p\left(\binom{n}{m}\right) = \sum_{j=1}^{t-1} \left( \left[\frac{n}{p^j}\right] - \left[\frac{m}{p^j}\right] - \left[\frac{n-m}{p^j}\right] \right)$
$$
= \sum_{j=1}^{t-1} 0 = 0.
$$

Therefore, $p \nmid \binom{n}{m}$.
(2) Necessity.

If $p \nmid \binom{n}{m}$, then $v_p\left(\binom{n}{m}\right) = 0$.
Assume there exists $i_0 \in \{1, 2, \cdots, t\}$ such that $a_{i_0} < b_{i_0}$ (it is easy to see that $i_0 \geq 2$). Then
$$
\left[\frac{n}{p^{t+1-i_0}}\right] - \left[\frac{m}{p^{t+1-i_0}}\right] - \left[\frac{n-m}{p^{t+1-i_0}}\right] \geq 1.
$$

Thus, $0 = v_p\left(\binom{n}{m}\right)$
$$
\begin{array}{l}
= \sum_{j=1}^{t-1} \left( \left[\frac{n}{p^j}\right] - \left[\frac{m}{p^j}\right] - \left[\frac{n-m}{p^j}\right] \right) \\
= \sum_{\substack{j=1 \\ j \neq i+1-i_0}}^{i-1} \left( \left[\frac{n}{p^j}\right] - \left[\frac{m}{p^j}\right] - \left[\frac{n-m}{p^j}\right] \right) + \\
\left[\frac{n}{p^{t+1-i_0}}\right] - \left[\frac{m}{p^{t+1-i_0}}\right] - \left[\frac{n-m}{p^{t+1-i_0}}\right] \\
\geq 1.
\end{array}
$$

Contradiction.
Back to the original problem.
Let $n = (q a_1 a_2 \cdots a_t)_p (q \neq 0)$ satisfy the condition.
Since $\binom{n}{0}, \binom{n}{1}, \cdots, \binom{n}{p-1}$ contain exactly $k$ numbers divisible by $p$, according to the lemma, we get $a_t = p - k - 1$.
If $a_1, a_2, \cdots, a_{t-1}$ are not all $p-1$, let
$$
i = \left( \underbrace{(p-1)(p-1) \cdots (p-1)}_{t-1 \text{ times}} 0 \right)_p,
$$

then according to the lemma, we know that $\binom{n}{i}, \binom{n}{i+1}, \cdots, \binom{n}{i+p-1}$ are all multiples of $p$, contradiction.
Therefore, $a_1 = a_2 = \cdots = a_{t-1} = p-1$.
Thus, $n = (q+1)p^t - k - 1 \left(1 \leq q \leq p-1, t \in \mathbb{Z}_+\right)$.
On the other hand, when $n = (q+1)p^t - k - 1$, using the lemma, it is easy to verify that $n$ satisfies the condition.
In summary,
$$
n = (q+1)p^t - k - 1 \left(1 \leq q \leq p-1, t \in \mathbb{Z}_+\right).
$$