4. (Full score: 50 points) Find all positive integers $k$ with the following property: for any positive integer $n$, $2^{(k-1) n+1}$ does not divide $\frac{(k n)!}{n!}$.


Solution: For a positive integer $m$, let $v_{2}(m)$ denote the power of the prime factor 2 in the standard factorization of the positive integer $m$. It is well known that $v_{2}(m!)=m-S(m)$, where $S(m)$ represents the sum of the digits of the positive integer $m$ in its binary representation.
Since $2^{(k-1) n+1}$ does not divide $\frac{(k n)!}{n!}$, this is equivalent to $v_{2}\left(\frac{(k n)!}{n!}\right) \leq(k-1) n$, i.e., $k n-v_{2}((k n)!) \geq n-v_{2}(n!)$, and by (1), the problem is equivalent to finding all positive integers $k$ such that $S(k n) \geq S(n)$ for any positive integer $n$. 10 points We prove that all such $k$ are $2^{a}(a=0,1,2, \cdots)$.
On one hand, since $S\left(2^{a} n\right)=S(n)$ for any positive integer $n$, $k=2^{a}$ satisfies the condition. 20 points On the other hand, if $k$ is not a power of 2, let $k=2^{a} \cdot q, a \geq 0, q$ be an odd number greater than 1.
Below, we construct a positive integer $n$ such that $S(k n)0$, so $\frac{2^{u}-1}{q} \cdot 2^{l u+a_{i}}(l=0,1, \cdots, t-1)$ do not contain 1 in their binary representations, hence by (2) we know $S\left(\frac{m}{q}\right)=1+S\left(\frac{2^{u}-1}{q}\right) \cdot t>t=S(m)$,
therefore the above selected $m$ meets the requirement.
Combining the two aspects above, the required $k$ is $2^{a}(a=0,1,2, \cdots) .50$ points