Four, (50 points) Find all positive integers $k$ with the following property: for any positive integer $n, 2^{(k-1) n+1} \frac{(k n)!}{n!}$.

保留源文本的换行和格式，翻译结果如下：

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Four, (50 points) Find all positive integers $k$ with the following property: for any positive integer $n, 2^{(k-1) n+1} \frac{(k n)!}{n!}$.
```
For a positive integer $m$, let $v_{2}(m)$ be the power of the prime factor 2 in the standard factorization of $m$. Then
$$
v_{2}(m!)=m-S(m),
$$

where $S(m)$ represents the sum of the digits of the positive integer $m$ in its binary representation.
From $2^{(k-1) n+1}\left\{\frac{(k n)!}{n!} \Leftrightarrow v_{2}\left(\frac{(k n)!}{n!}\right) \leqslant(k-1) n\right.$
$$
\Rightarrow k n-v_{2}((k n)!) \geqslant n-v_{2}(n!) \text {. }
$$

Thus, by equation (1), the problem is equivalent to finding all positive integers $k$ such that for any positive integer $n$, we have
$$
S(k n) \geqslant S(n) \text {. }
$$

Next, we prove that all such $k$ are $2^{a}$ $(a=0,1, \cdots)$.
On one hand, because for any positive integer $n$, we have
$$
S\left(2^{a} n\right)=S(n),
$$

so $k=2^{a}$ satisfies the condition.
On the other hand, if $k$ is not a power of 2, let $k=2^{a} q$ (where $a \geqslant 0, q$ is an odd number greater than 1).
We construct a positive integer $n$ such that
$S(k n)0 \text {, then } \frac{2^{u}-1}{q} \cdot 2^{l u+\alpha_{t}}(l=0,1, \cdots, t-1)
$$

do not contain 1 in their binary representation.
Therefore, by equation (2), we have
$$
S\left(\frac{m}{q}\right)=1+S\left(\frac{2^{u}-1}{q}\right) t>t=S(m) \text {. }
$$

Thus, the above chosen $m$ meets the requirement.
In summary, the required $k$ is $2^{a}(a=0,1, \cdots)$.