$ 1000 \le \overline{abcd} < 10000$
$ 91 \le (a + b + c + d)^2 \le 909$
$ (a + b + c + d) \in \{10,11,\cdots, 30\}$
On the other hand, $ x \equiv \overline{abcd} \equiv a + b + c + d (9) \Rightarrow x \equiv 2 x^2 (9) \Rightarrow x \equiv 0 (9)$ or $ x \equiv 5 (9)$
Combining both implies $ (a + b + c + d) \in \{14,18,23,27\}$
obtaining the following solutions for $ \overline{abcd} \in \{2156,3564,5819\}$