Four, (16 points) Let the four-digit number $\overline{a b c d}$ be a perfect square, and $\overline{a b}=2 \overline{c d}+1$. Find this four-digit number.
Let $\overline{a b c d}=m^{2}$, then $32 \leqslant m \leqslant 99$.
Suppose $\overline{c d}=x$, then $\overline{a b}=2 x+1$. Therefore,
$$
100(2 x+1)+x=m^{2} \text {, }
$$

which simplifies to $67 \times 3 x=(m+10)(m-10)$.
Since 67 is a prime number, at least one of $m+10$ and $m-10$ must be a multiple of 67.
(1) If $m+10=67 k$ (where $k$ is a positive integer), because $32 \leqslant m \leqslant 99$, then $m+10=67$, i.e., $m=57$.

Upon verification, $57^{2}=3249$, which does not meet the condition, so it is discarded.
(2) If $m-10=67 k$ (where $k$ is a positive integer), then $m-10=67, m=77$.
Thus, $\overline{a b c d}=77^{2}=5929$.