Find all positive integers $k$ such that for any positive integer $n$, $2^{(k-1)n+1}$ does not divide $\frac{(kn)!}{n!}$.
1. **Define the valuation function and use Legendre's formula:**

   Let \( v_p(x) \) denote the highest power of a prime \( p \) that divides \( x \). According to Legendre's formula, for a prime \( p \),
   \[
   v_p(n!) = \frac{n - S_p(n)}{p-1}
   \]
   where \( S_p(n) \) is the sum of the digits of \( n \) when written in base \( p \).

2. **Apply Legendre's formula to the given expression:**

   We need to find \( v_2 \left( \frac{(kn)!}{n!} \right) \). Using Legendre's formula, we get:
   \[
   v_2 \left( \frac{(kn)!}{n!} \right) = v_2((kn)!) - v_2(n!)
   \]
   Applying Legendre's formula to each factorial, we have:
   \[
   v_2((kn)!) = \frac{kn - S_2(kn)}{1} = kn - S_2(kn)
   \]
   \[
   v_2(n!) = \frac{n - S_2(n)}{1} = n - S_2(n)
   \]
   Therefore,
   \[
   v_2 \left( \frac{(kn)!}{n!} \right) = (kn - S_2(kn)) - (n - S_2(n)) = (k-1)n + S_2(n) - S_2(kn)
   \]

3. **Condition for divisibility:**

   For \( 2^{(k-1)n+1} \) to not divide \( \frac{(kn)!}{n!} \), we need:
   \[
   v_2 \left( \frac{(kn)!}{n!} \right) < (k-1)n + 1
   \]
   Substituting the expression from step 2, we get:
   \[
   (k-1)n + S_2(n) - S_2(kn) < (k-1)n + 1
   \]
   Simplifying, we find:
   \[
   S_2(n) - S_2(kn) < 1
   \]
   This implies:
   \[
   S_2(kn) \geq S_2(n)
   \]

4. **Analyze the condition \( S_2(kn) \geq S_2(n) \):**

   - If \( k = 1 \), this condition is trivially true because \( S_2(n) = S_2(n) \).
   - Notice that if \( k \) satisfies the condition, then \( 2k \) also satisfies it. This is because multiplying by 2 shifts the binary digits to the left, which does not decrease the number of 1s.

5. **Check for odd \( k \):**

   We need to show that there is no odd \( k > 1 \) such that \( 2^{(k-1)n+1} \) does not divide \( \frac{(kn)!}{n!} \) for all \( n \).

   - Let \( m \) be a multiple of \( k \) with a minimal number of 1s in its binary representation (suppose it has \( v \) many 1s) and take \( m \) to be odd.
   - Let \( m = 2^a + b \) with \( 2^a > b \).
   - Consider the number \( w = 2^{a+s\varphi(k)} + b \), which is also a multiple of \( k \) with \( v \) ones for every \( s \in \mathbb{Z}^+ \).

6. **Prove that \( w/k \) must have more than \( v \) digits 1 for sufficiently large \( s \):**

   - Assume otherwise, and pick \( s \) so that \( w/k \) has at least \( z \) consecutive zeroes, where \( z \) is the length of the binary representation of \( k \).
   - For sufficiently large \( s \), the number of digits 1 in the binary representation of \( w/k \cdot k \) is greater than \( v \), leading to a contradiction.

Thus, the only possible values for \( k \) are powers of 2.

The final answer is the set of all powers of \( \boxed{ 2 } \).