China Girls' Mathematical Olympiad

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 $A_i\ne A_{i+1}$, 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 $$A_1=\{1\},A_2=\{1,2\},A_3=\{2\},A_4=\{2,3\},\dots ,A_{2n-2}=\{n-1,n\},A_{2n-1}=\{n\},A_{2n}=\{n,1\}.$$