It is easy to see that $A$ can not contain more than two numbers of any parity. Combined with $n \ge 4$, this forces $A$ to have exactly $4$ elements, two of each parity. The difference between the two even (odd) numbers must be $2$, therefore we can have two types of sets: $\{2013, 2015, 2k, 2k + 2\}$ and $\{2015, 2017, 2k, 2k + 2\}$.

In the first case, the differences $|(2k + 2) - 2013|$, $|2k - 2013|$, and $|2k - 2015|$ give different remainders upon division by $3$, hence one of these differences has to be $3$. Checking all the possibilities leads to the solutions $A = \{2013, 2015, 2008, 2010\}$ and $A = \{2013, 2015, 2018, 2020\}$.

In the second case, one of the differences $|(2k + 2) - 2015|$, $|2k - 2015|$, and $|2k - 2017|$ has to be $3$. Studying the cases we get the answers $\{2015, 2017, 2010, 2012\}$, $\{2015, 2017, 2020, 2022\}$.

In conclusion
$$\begin{aligned} A_1 &= \{2008, 2010, 2013, 2015\}, & A_2 &= \{2010, 2012, 2015, 2017\}, \\ A_3 &= \{2013, 2015, 2018, 2020\}, & A_4 &= \{2015, 2017, 2020, 2022\}. \end{aligned}$$