The answer is $n$.

We consider each summand $s_i = \frac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|}$.
If $A_i \cap A_{i+1}$ is the empty set, then $s_i = 0$.
If $A_i \cap A_{i+1}$ is nonempty, because $A_i \neq A_{i+1}$, at least one of $A_i$ and $A_{i+1}$ has more than one element, that is, $\max\{|A_i|, |A_{i+1}|\} \ge 2$. Because $A_i \cap A_{i+1}$ is a subset of each of $A_i$ and $A_{i+1}$, $|A_i \cap A_{i+1}| \le \min\{|A_i|, |A_{i+1}|\}$ and
$$s_i = \frac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|} \le \frac{\min\{|A_i|, |A_{i+1}|\}}{\max\{|A_i|, |A_{i+1}|\} \cdot \min\{|A_i|, |A_{i+1}|\}} \le \frac{1}{2}.$$
It follows that
$$\sum_{i=1}^{2n} \frac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|} \le \sum_{i=1}^{2n} \frac{1}{2} = n.$$
This upper bound can be achieved with sets
$$A_1 = \{1\}, A_2 = \{1, 2\}, A_3 = \{2\}, A_4 = \{2, 3\}, \dots, A_{2n-2} = \{n-1, n\}, A_{2n-1} = \{n\}, A_{2n} = \{n, 1\}.$$