For every sequence $ (x_1, x_2, \ldots, x_n)$ of non-zero natural prime numbers, $ \{1, 2, \ldots, n\}$ arranged in any order, denote by $ f(s)$ the sum of absolute values of the differences between two consecutive members of $ s.$ Find the maximum value of $ f(s)$ where $ s$ runs through the set of all such sequences, i.e. for all sequences $ s$ with the given properties.
1. **Understanding the Problem:**
   We are given a sequence \( s = (x_1, x_2, \ldots, x_n) \) where \( x_i \) are distinct natural numbers from the set \(\{1, 2, \ldots, n\}\). We need to find the maximum value of \( f(s) \), where \( f(s) \) is defined as the minimum absolute difference between consecutive elements in the sequence \( s \).

2. **Rephrasing the Problem:**
   We need to compute:
   \[
   \max_{\sigma \in \mathcal{S}_n} \min_{1 \leq k \leq n-1} |\sigma(k+1) - \sigma(k)|
   \]
   where \(\mathcal{S}_n\) is the set of all permutations of \(\{1, 2, \ldots, n\}\).

3. **Case Analysis:**
   - **For \( n = 2m \) (even):**
     Consider the sequence \(\sigma = (m+1, 1, m+2, 2, \ldots, 2m, m)\). The differences between consecutive terms are:
     \[
     |(m+1) - 1| = m, \quad |1 - (m+2)| = m+1, \quad |(m+2) - 2| = m, \quad \ldots, \quad |2m - m| = m
     \]
     The minimum difference in this sequence is \( m \). We need to show that no other sequence can have a minimum difference greater than \( m \).

     Since \( m \) must appear in any sequence, the maximum possible difference involving \( m \) is \( m \) (i.e., \( |m - \ell| \leq m \) for all \( 1 \leq \ell \leq 2m \)).

   - **For \( n = 2m+1 \) (odd):**
     Consider the sequence \(\sigma = (m+1, 1, m+2, 2, \ldots, 2m, m, 2m+1)\). The differences between consecutive terms are:
     \[
     |(m+1) - 1| = m, \quad |1 - (m+2)| = m+1, \quad |(m+2) - 2| = m, \quad \ldots, \quad |2m - m| = m, \quad |m - (2m+1)| = m+1
     \]
     The minimum difference in this sequence is \( m \). We need to show that no other sequence can have a minimum difference greater than \( m \).

     Since \( m+1 \) must appear in any sequence, the maximum possible difference involving \( m+1 \) is \( m \) (i.e., \( |m+1 - \ell| \leq m \) for all \( 1 \leq \ell \leq 2m+1 \)).

4. **Conclusion:**
   For both even and odd \( n \), the maximum value of \( f(s) \) is \( \lfloor n/2 \rfloor \).

The final answer is \( \boxed{ \lfloor n/2 \rfloor } \).