Which positive integers $m$ are such that $k^m - 1$ is divisible by $2^m$ for all odd numbers $k \ge 3$?

To determine which positive integers \( m \) satisfy the condition that \( 2^m \) divides \( k^m - 1 \) for all odd numbers \( k \ge 3 \), we can use the Lifting the Exponent (LTE) lemma. Let's go through the steps in detail:

1. **Apply the Lifting the Exponent Lemma:**
   The LTE lemma states that for odd integers \( k \) and \( n \), and for \( p = 2 \), we have:
   \[
   v_2(k^n - 1) = v_2(k - 1) + v_2(k + 1) + v_2(n) - 1
   \]
   where \( v_2(x) \) denotes the highest power of 2 dividing \( x \).

2. **Evaluate \( v_2(k^m - 1) \):**
   Since \( k \) is an odd number, \( k - 1 \) and \( k + 1 \) are consecutive even numbers. Therefore, one of them is divisible by 2 but not by 4, and the other is divisible by 4. Hence:
   \[
   v_2(k - 1) + v_2(k + 1) = 1 + 2 = 3
   \]
   Thus, applying the LTE lemma:
   \[
   v_2(k^m - 1) = 3 + v_2(m) - 1 = 2 + v_2(m)
   \]

3. **Determine the condition for divisibility:**
   For \( 2^m \) to divide \( k^m - 1 \), we need:
   \[
   m \leq v_2(k^m - 1) = 2 + v_2(m)
   \]

4. **Analyze the inequality \( m \leq 2 + v_2(m) \):**
   - If \( m = 1 \), then \( v_2(1) = 0 \), and \( 1 \leq 2 + 0 \) is true.
   - If \( m = 2 \), then \( v_2(2) = 1 \), and \( 2 \leq 2 + 1 \) is true.
   - If \( m = 4 \), then \( v_2(4) = 2 \), and \( 4 \leq 2 + 2 \) is true.
   - For \( m = 8 \), \( v_2(8) = 3 \), and \( 8 \leq 2 + 3 \) is false.
   - For \( m = 3 \), \( v_2(3) = 0 \), and \( 3 \leq 2 + 0 \) is false.
   - For \( m = 5 \), \( v_2(5) = 0 \), and \( 5 \leq 2 + 0 \) is false.
   - For \( m = 6 \), \( v_2(6) = 1 \), and \( 6 \leq 2 + 1 \) is false.

   Therefore, the only values of \( m \) that satisfy the inequality are \( m = 1, 2, 4 \).

The final answer is \( \boxed{1, 2, 4} \).