Find all positive integers $k$ with the following property: for any positive integer $n, 2^{(k-1) n+1} \left\lvert\, \frac{(k n)!}{n!}\right.$.
【Analysis】From question 1(1), we get $\operatorname{ord}_{2} n!=n-S_{2}(n)$.
Combining with the problem, we only need to consider the inequality
$$
\begin{array}{l}
\operatorname{ord}_{2} \frac{(k n)!}{n!} \\
=k n-S_{2}(k n)-\left(n-S_{2}(n)\right) \\
=(k-1) n+S_{2}(n)-S_{2}(k n) \\
\leqslant(k-1) n
\end{array}
$$

for any positive integer $n$.
This inequality can be simplified to
$$
S_{2}(n) \leqslant S_{2}(k n) \text {. }
$$

First, we prove that $k=2^{s}(s \in \mathbf{N})$ satisfies the problem's conditions.
When $k=2^{s}(s \in \mathbf{N})$, in binary, $k n$ is equivalent to appending $s$ zeros to $n$, i.e.,
$$
S_{2}(k n)=S_{2}(n) \text {. }
$$

Then, for any positive integer $n, 2^{(k-1) n+1} \frac{(k n)!}{n!}$.
When $k \neq 2^{s}$, assume
$k=2^{a} p(a>0$, and $p$ is an odd number greater than 1 $)$.
Below, we construct a positive integer $n$ such that
$$
S_{2}(k n)1\right) \text {. }
$$

Take $m=\sum_{i=1}^{t-1} 2^{\alpha_{i}}+2^{\alpha_{t}+t u}$.
Then $S_{2}(m)=t$, and
$$
\begin{array}{l}
m=p+2^{\alpha_{t}}\left(2^{t u}-1\right) \equiv 0(\bmod p) . \\
\text { Hence } \frac{m}{p}=1+2^{\alpha_{i}} \cdot \frac{2^{t u}-1}{p} \\
=1+2^{\alpha_{t}} \cdot \frac{2^{u}-1}{p} \sum_{i=0}^{t-1} 2^{i u} \\
=1+\sum_{i=0}^{t-1}\left(\frac{2^{u}-1}{p} \cdot 2^{i u+\alpha_{i}}\right) .
\end{array}
$$

Since the highest power in the binary representation of the positive integer $\frac{2^{u}-1}{p}$ is less than $u$, for any integers $i 、 j(0 \leqslant i0 \text {, then } \frac{2^{u}-1}{p} \cdot 2^{i u+\alpha_{t}}(i=0,1, \cdots, t-1)
$$

does not contain 1 in its binary representation. Therefore,
$$
S_{2}\left(\frac{m}{p}\right)=1+S_{2}\left(\frac{2^{u}-1}{p}\right) t>t=S_{2}(m) .
$$

Taking $n=m$ satisfies the requirement.
In summary, the values of $k$ that meet the conditions are $2^{s}(s=0,1$, $\cdots$.

Similarly, the fourth question of the 2015 Tsinghua University Mathematics Experience Camp can also be proven using part (1) of question 1.