Romanian Mathematical Olympiad

Since $\sqrt{\overline{ab} - 4} + \sqrt{\overline{cd}} < 10 + 10 = 20$, we obtain $p = 2$ or $p = 3$.
If $p = 2$, then $\overline{cd} = \overline{ab} + 4$ and $\sqrt{\overline{ab} - 4} + \sqrt{\overline{ab} + 4} = 4$, therefore $\overline{ab} = 5$, false.
If $p = 3$, then $\overline{cd} = \overline{ab} + 5$ and $\sqrt{\overline{ab} - 4} + \sqrt{\overline{ab} + 5} = 9$. We obtain the unique solution $\overline{ab} = 20$, thus $\overline{abcd} = 2025$.

Alternative solution:
Since $\overline{ab}-4$, $\overline{cd}$ and $\sqrt{\overline{ab}-4}+\sqrt{\overline{cd}}$ are integers, the natural numbers $k, n$ exist, such that $\overline{ab}-4 = k^2$ and $\overline{cd} = n^2$, hence $k+n = p^2$. (1)
We obtain $p+6 = \overline{cd} - \overline{ab} + 4 = n^2 - k^2 = (n-k)(n+k)$. Using (1), we deduce that $p+6 = (n-k) \cdot p^2$, therefore $p \mid (p+6)$, so $p \in \{2, 3\}$.
If $p = 2$, we obtain $k = 1$ and $n = 3$, therefore $\overline{ab} = 5$, which is false. If $p = 3$, we obtain $k = 4$ and $n = 5$, so $\overline{ab} = 20$ and $\overline{cd} = 25$, which satisfies the hypothesis. Therefore, the solution is 2025.