Let $n$ be an integer greater than two, and let $A_1,A_2, \cdots , A_{2n}$ be pairwise distinct subsets of $\{1, 2, ,n\}$. Determine the maximum value of
\[\sum_{i=1}^{2n} \dfrac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|}\]
Where $A_{2n+1}=A_1$ and $|X|$ denote the number of elements in $X.$
1. **Initial Setup and Definitions:**
   Let \( n \) be an integer greater than two, and let \( A_1, A_2, \ldots, A_{2n} \) be pairwise distinct subsets of \( \{1, 2, \ldots, n\} \). We need to determine the maximum value of the sum:
   \[
   \sum_{i=1}^{2n} \frac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|}
   \]
   where \( A_{2n+1} = A_1 \) and \( |X| \) denotes the number of elements in \( X \).

2. **Non-Empty Subsets:**
   Note that \( |A_j| \neq 0 \) for all \( j \in \{1, 2, \ldots, 2n\} \) because the subsets are pairwise distinct and non-empty.

3. **Upper Bound Using Intersection Property:**
   For any two subsets \( X \) and \( Y \) of a set \( S \), we have:
   \[
   |X \cap Y| \leq \min\{|X|, |Y|\}
   \]
   Therefore,
   \[
   \frac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|} \leq \frac{\min\{|A_i|, |A_{i+1}|\}}{|A_i| \cdot |A_{i+1}|}
   \]

4. **Defining \( B_i \):**
   Let \( B_i = \max\{|A_i|, |A_{i+1}|\} \). Then,
   \[
   \frac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|} \leq \frac{1}{B_i}
   \]

5. **Summing the Inequalities:**
   Summing over all \( i \) from 1 to \( 2n \), we get:
   \[
   \sum_{i=1}^{2n} \frac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|} \leq \sum_{i=1}^{2n} \frac{1}{B_i}
   \]

6. **Lower Bound for \( B_i \):**
   We claim that \( |B_i| \geq 2 \) for all \( i \in \{1, 2, \ldots, 2n\} \). Suppose there exists a \( j \) such that \( |B_j| = 1 \). This implies \( |A_j| = 1 \) and \( |A_{j+1}| = 1 \). Since \( A_j \) and \( A_{j+1} \) are distinct, \( A_j \cap A_{j+1} = \emptyset \), making the term \( \frac{|A_j \cap A_{j+1}|}{|A_j| \cdot |A_{j+1}|} = 0 \), which decreases the sum.

7. **Minimum Value for \( B_i \):**
   Therefore, \( |B_i| \geq 2 \) for all \( i \). This gives:
   \[
   \sum_{i=1}^{2n} \frac{1}{B_i} \leq \sum_{i=1}^{2n} \frac{1}{2} = n
   \]

8. **Construction for Maximum Value:**
   To achieve the maximum value \( n \), consider the construction:
   \[
   A_{2i-1} = \{2i-1\}, \quad A_{2i} = \{2i-1, 2i\}
   \]
   For this construction:
   \[
   \frac{|A_{2i-1} \cap A_{2i}|}{|A_{2i-1}| \cdot |A_{2i}|} = \frac{1}{1 \cdot 2} = \frac{1}{2}
   \]
   Summing over all \( i \):
   \[
   \sum_{i=1}^{2n} \frac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|} = 2n \cdot \frac{1}{2} = n
   \]

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