4. Find all positive integers $k$ such that for any positive integer $n, 2^{(k-1) n+1}$ does not divide $\frac{(k n)!}{n!}$.
For a positive integer $m$, let $v_{2}(m)$ denote the exponent 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. $2^{(k-1) n+1}$ does not divide $\frac{(k n)!}{n!}$ is equivalent to $v_{2}\left(\frac{(k n)!}{n!}\right) \leqslant(k-1) n$ $\Leftrightarrow[k n-S(k n)]-[n-S(n)] \leqslant(k-1) n \Leftrightarrow S(k n) \geqslant S(n)$ for any positive integer $n$. We prove: all $k$ that satisfy the condition are $2^{a}(a=0,1,2, \cdots)$. On one hand, since $S\left(2^{a} n\right)=S(n)$ for any positive integer $n$, i.e., $k=2^{a}$ satisfies the condition; on the other hand, if $k$ is not a power of 2, let $k=2^{a} \cdot q, a \geqslant 0, q$ be an odd number greater than 1. Below, we construct a positive integer $n$ such that $S(k n)t=S(m)$ meets the requirement. In summary, the required $k=2^{a}(a=0,1,2, \cdots)$.