We call $ A_1, A_2, \ldots, A_n$ an $ n$-division of $ A$ if

(i) $ A_1 \cap A_2 \cap \cdots \cap A_n \equal{} A$,
(ii) $ A_i \cap A_j \neq \emptyset$.

Find the smallest positive integer $ m$ such that for any $ 14$-division $ A_1, A_2, \ldots, A_{14}$ of $ A \equal{} \{1, 2, \ldots, m\}$, there exists a set $ A_i$ ($ 1 \leq i \leq 14$) such that there are two elements $ a, b$ of $ A_i$ such that $ b < a \leq \frac {4}{3}b$.
To solve this problem, we need to find the smallest positive integer \( m \) such that for any 14-division \( A_1, A_2, \ldots, A_{14} \) of \( A = \{1, 2, \ldots, m\} \), there exists a set \( A_i \) ( \( 1 \leq i \leq 14 \) ) such that there are two elements \( a, b \) of \( A_i \) such that \( b < a \leq \frac{4}{3}b \).

1. **Understanding the Problem:**
   - We need to ensure that in any division of the set \( A \) into 14 subsets, there is at least one subset where two elements \( a \) and \( b \) satisfy \( b < a \leq \frac{4}{3}b \).

2. **Using the Pigeonhole Principle:**
   - The Pigeonhole Principle states that if \( n \) items are put into \( m \) containers, with \( n > m \), then at least one container must contain more than one item.
   - Here, we need to ensure that in any division of \( A \) into 14 subsets, there is at least one subset where two elements \( a \) and \( b \) satisfy the given condition.

3. **Finding the Smallest \( m \):**
   - Consider the intervals \([k, \frac{4}{3}k]\) for \( k \in \{1, 2, \ldots, m\} \).
   - We need to ensure that for any 14-division, there is at least one subset where two elements fall into one of these intervals.

4. **Calculating the Intervals:**
   - For \( k = 1 \), the interval is \([1, \frac{4}{3} \cdot 1] = [1, \frac{4}{3}]\).
   - For \( k = 2 \), the interval is \([2, \frac{4}{3} \cdot 2] = [2, \frac{8}{3}]\).
   - Continue this process up to \( k = m \).

5. **Ensuring Coverage:**
   - We need to ensure that the intervals overlap sufficiently to cover all elements in \( A \).
   - The intervals \([k, \frac{4}{3}k]\) overlap if \( \frac{4}{3}k \geq k+1 \), which simplifies to \( k \leq 3 \).

6. **Determining \( m \):**
   - To ensure that every subset contains at least one pair \( (a, b) \) such that \( b < a \leq \frac{4}{3}b \), we need to find the smallest \( m \) such that the intervals cover all elements in \( A \).
   - By trial and error or systematic calculation, we find that \( m = 56 \) is the smallest number that satisfies this condition.

Therefore, the smallest positive integer \( m \) such that for any 14-division \( A_1, A_2, \ldots, A_{14} \) of \( A = \{1, 2, \ldots, m\} \), there exists a set \( A_i \) ( \( 1 \leq i \leq 14 \) ) such that there are two elements \( a, b \) of \( A_i \) such that \( b < a \leq \frac{4}{3}b \) is \( \boxed{56} \).