Example 1 Find all positive integers $k$ with the following property: for any positive integer $n$, there is $2^{(k-1) n+1} \times \frac{(k n)!}{n!}$.
(2015, National High School Mathematics League)
From Conclusion 1, we know that the power of 2 in $\frac{(k n)!}{n!}$ is
$$
\begin{array}{l}
k n-s(k n)-(n-s(n)) \\
=(k-1) n-(s(k n)-s(n)) .
\end{array}
$$

Then, the required $k$ is such that for any positive integer $n$, we have $s(k n) \geqslant s(n)$.
For all $k$ being powers of 2, we have
$$
s(k n)=s\left(2^{m} n\right)=s(n) \text {, }
$$

which satisfies the requirement.
For $k$ that are not powers of 2, since multiplying by powers of 2 does not change the sum of the binary digits of a number, we can assume $k$ is an odd number greater than 1.
By Euler's theorem, we know $k \mid\left(2^{\varphi(k)}-1\right)$.
We can set $k t=2^{m}-1$. Then $t$ has at most $m-1$ binary digits.
Let $n=t\left(3+\sum_{i=1}^{q} 2^{m i+1}\right)$. Then
$$
k n=k t\left(3+\sum_{i=1}^{q} 2^{m i+1}\right)=2^{(q+1) m+1}+2^{m}-3 \text {. }
$$

Thus, $s(k n)=m, s(n)>q$.
Taking $q$ as $m+1$, we find that there exists $n$ such that
$$
s(k n)<s(n) \text {. }
$$

In summary, the required $k$ is $2^{a}(a=0,1, \cdots)$.