What is the least positive integer $m$ such that the following is true?

[i]Given $\it m$ integers between $\it1$ and $\it{2023},$ inclusive, there must exist two of them $\it a, b$ such that $1 < \frac ab \le 2.$  [/i] \[\mathrm a. ~ 10\qquad \mathrm b.~11\qquad \mathrm c. ~12 \qquad \mathrm d. ~13 \qquad \mathrm e. ~1415\]
To solve this problem, we need to find the smallest positive integer \( m \) such that among any \( m \) integers chosen from the set \(\{1, 2, \ldots, 2023\}\), there exist two integers \( a \) and \( b \) satisfying \( 1 < \frac{a}{b} \leq 2 \).

1. **Understanding the Problem:**
   - We need to ensure that for any \( m \) integers chosen from \(\{1, 2, \ldots, 2023\}\), there are two integers \( a \) and \( b \) such that \( 1 < \frac{a}{b} \leq 2 \).
   - This means \( a \) must be at most twice \( b \) but greater than \( b \).

2. **Constructing a Sequence:**
   - To avoid having \( 1 < \frac{a}{b} \leq 2 \), we need to construct a sequence where each number is more than twice the previous number.
   - Start with the smallest number, say \( a_1 = 1 \). The next number \( a_2 \) must be greater than \( 2 \times 1 = 2 \), so \( a_2 \geq 3 \).
   - Continue this process: \( a_3 \geq 2 \times 3 = 6 \), \( a_4 \geq 2 \times 6 = 12 \), and so on.

3. **Finding the Maximum Sequence:**
   - We need to find the maximum number of terms in this sequence that can fit within the range \(\{1, 2, \ldots, 2023\}\).
   - Start with \( a_1 = 1 \):
     \[
     \begin{align*}
     a_1 &= 1, \\
     a_2 &= 3, \\
     a_3 &= 6, \\
     a_4 &= 12, \\
     a_5 &= 24, \\
     a_6 &= 48, \\
     a_7 &= 96, \\
     a_8 &= 192, \\
     a_9 &= 384, \\
     a_{10} &= 768, \\
     a_{11} &= 1536.
     \end{align*}
     \]
   - The next term would be \( a_{12} = 2 \times 1536 = 3072 \), which is outside the range \(\{1, 2, \ldots, 2023\}\).

4. **Conclusion:**
   - We can have at most 11 terms in this sequence without violating the condition \( 1 < \frac{a}{b} \leq 2 \).
   - Therefore, if we choose 12 numbers, by the pigeonhole principle, there must be at least two numbers \( a \) and \( b \) such that \( 1 < \frac{a}{b} \leq 2 \).

The final answer is \(\boxed{12}\).