For given integer $n \geq 3$, set $S =\{p_1, p_2, \cdots, p_m\}$ consists of permutations $p_i$ of $(1, 2, \cdots, n)$. Suppose that among every three distinct numbers in $\{1, 2, \cdots, n\}$, one of these number does not lie in between the other two numbers in every permutations $p_i$ ($1 \leq i \leq m$). (For example, in the permutation $(1, 3, 2, 4)$, $3$ lies in between $1$ and $4$, and $4$ does not lie in between $1$ and $2$.) Determine the maximum value of $m$.
1. **Define the problem and base case:**
   We need to determine the maximum value of \( m \) such that the set \( S = \{p_1, p_2, \cdots, p_m\} \) of permutations of \((1, 2, \cdots, n)\) satisfies the given condition. We denote this maximum value by \( a_n \). We aim to show that \( a_n = 2^{n-1} \) by induction.

   For the base case \( n = 3 \), we need to show that \( a_3 = 2^{3-1} = 4 \). Consider the permutations of \(\{1, 2, 3\}\):
   \[
   \{(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)\}
   \]
   We can verify that any set of 4 permutations from these 6 will satisfy the condition. Thus, \( a_3 = 4 \).

2. **Inductive hypothesis:**
   Assume that for some \( n \geq 3 \), \( a_n = 2^{n-1} \). We need to show that \( a_{n+1} = 2^n \).

3. **Constructing permutations for \( n+1 \):**
   Clearly, \( a_{n+1} \geq 2a_n \) because we can construct permutations for \( n+1 \) by adding \( n+1 \) either at the front or the end of each permutation in a good set of size \( a_n \). This gives us \( 2a_n = 2 \cdot 2^{n-1} = 2^n \).

4. **Contradiction argument:**
   Assume for contradiction that \( a_{n+1} \geq 2a_n + 1 = 2^n + 1 \). Let \( \{p_1, p_2, \cdots, p_{2^n+1}\} \) be such a good set. Consider the permutations \( p_i \) without the element \( n+1 \). These restricted permutations form a good set for \(\{1, 2, \cdots, n\}\), and by the Pigeonhole Principle, at least three of the \( p_i \), say \( p_1, p_2, p_3 \), must map to the same restricted permutation of \(\{1, 2, \cdots, n\}\).

5. **Analyzing the positions of \( n+1 \):**
   Since \( p_1, p_2, p_3 \) are distinct, the positions of \( n+1 \) in these permutations must be different. Without loss of generality, assume the position of \( n+1 \) in \( p_1 \) is further left than in \( p_2 \), which is further left than in \( p_3 \). Let \( a \) be an element to the right of \( n+1 \) in \( p_1 \), and \( b \) be an element to the right of \( n+1 \) in \( p_2 \). Clearly, \( a \) and \( b \) are distinct.

6. **Violation of the condition:**
   Consider the triple \(\{n+1, a, b\}\). In \( p_1 \), \( n+1 \) is to the left of \( a \), and in \( p_2 \), \( n+1 \) is to the left of \( b \). This implies that in \( p_3 \), \( n+1 \) is to the right of both \( a \) and \( b \), violating the condition that one of these numbers does not lie between the other two in every permutation.

7. **Conclusion:**
   Hence, our assumption that \( a_{n+1} \geq 2^n + 1 \) leads to a contradiction. Therefore, \( a_{n+1} = 2^n \), completing the induction.

\(\blacksquare\)

The final answer is \( \boxed{ 2^{n-1} } \)