Let the four-digit number $\overline{a b c d}$ be a perfect square, and its arithmetic square root can be expressed as $\sqrt{\overline{a b c d}}=\overline{a b}+\sqrt{\overline{c d}}$. How many such four-digit numbers are there?
Given $\sqrt{a b c d}=\overline{a b}+\sqrt{\overline{c d}}$, we know
$$
\begin{array}{l}
(\overline{a b})^{2}+2 \overline{a b} \cdot \sqrt{c d}+\overline{c d} \\
=\overline{a b c d}=100 \overline{a b}+\overline{c d} .
\end{array}
$$

Thus, $(\overline{a b})^{2}+2 \overline{a b} \cdot \sqrt{c d}=100 \overline{a b}$, which means
$$
\overline{a b}+2 \sqrt{c d}=100 \text {. }
$$

From equation (1), we know that $\overline{c d}$ is a perfect square. Therefore, $c d$ can be $01, 04, 09, 16, 25, 36, 49, 64, 81$, a total of 9 values.
From equation (1), the corresponding $\overline{a b}$ are
$$
98, 96, 94, 92, 90, 88, 86, 84, 82 \text {. }
$$

Thus, the four-digit numbers
$$
\begin{array}{l}
9801, 9604, 9409, 9216, 9025, \\
8836, 8649, 8464, 8281
\end{array}
$$

all satisfy the condition for $\overline{a b c d}$.
Rewriting equation (1) as
$$
\overline{a b}+\sqrt{\overline{c d}}=100-\sqrt{c d} .
$$

In fact,
$$
\begin{array}{l}
\sqrt{8281}=82+\sqrt{81}=100-\sqrt{81} ; \\
\sqrt{8464}=84+\sqrt{64}=100-\sqrt{64} ; \\
\sqrt{8649}=86+\sqrt{49}=100-\sqrt{49} ; \\
\sqrt{8836}=88+\sqrt{36}=100-\sqrt{36} ; \\
\sqrt{9025}=90+\sqrt{25}=100-\sqrt{25} ; \\
\sqrt{9216}=92+\sqrt{16}=100-\sqrt{16} ; \\
\sqrt{9409}=94+\sqrt{09}=100-\sqrt{09} ; \\
\sqrt{9604}=96+\sqrt{04}=100-\sqrt{04} ; \\
\sqrt{9801}=98+\sqrt{01}=100-\sqrt{01} .
\end{array}
$$

In summary, there are 9 four-digit numbers $\overline{a b c d}$ that meet the condition.