Let $\overline{a b c d}, \overline{d b c a}$ satisfy the conditions, and $a \leqslant d$. Then the sum
$$
\overline{a b c d}+\overline{d b c a}=1001 \cdot(a+d)+20(10 b+c)
$$

is a four-digit number, so $2 \leqslant a+d \leqslant 9$. The sum is also a perfect square $x^{2}$, and its last digit $a+d$ can only be $4,5,6,9$.
If $a+d=4$, then $\left(\frac{x}{2}\right)^{2}=1001+5(10 b+c)$, and the last digit is 1 or 6, so the last digit of $\frac{x}{2}$ is $1,4,6$ or 9. Noting that $1001 \leqslant\left(\frac{x}{2}\right)^{2}<1501$, i.e., $32 \leqslant \frac{x}{2} \leqslant 38$, so $\frac{x}{2}=34$ or 36. Thus, $x^{2}=68^{2}=4624=1313+3311=2312+2312$,
or $x^{2}=72^{2}=5184=1593+3591=2592+2592$.
If $a+d=5$, then $x^{2}=5(1001+40 b+4 c)$, so $5 \mid 4 c+1$. Thus, $c=1$ or 6. When $c=1$, $x^{2}=5^{2}(201+8 b)$. Hence, $201+8 b$ is a perfect square of an odd number, and lies between 201 and 273, so $201+8 b=225=15^{2}$. At this time, $x^{2}=75^{2}=5625=1314+4311=2313+3312$.
When $c=6$, $x^{2}=5^{2}(205+8 b)$. But $205+8 b$ leaves a remainder of 5 when divided by 8, so it is not a perfect square. Hence, $c \neq 6$.

If $a+d=6$, then $x^{2}$ is even, but not a multiple of 4. This is impossible.
If $a+d=9$, similarly, $x=97$, at this time
$$
\begin{aligned}
x^{2} & =97^{2}=9409=1208+8201=2207+7202 \\
& =3206+6203=4205+5204 .
\end{aligned}
$$

Therefore, there are 10 pairs of numbers that satisfy the given conditions.
Consider a pair of four-digit numbers, where swapping the first and last digits of one number results in the other (e.g., 1234 and 4231). How many such pairs of numbers have a sum that is a four-digit square number?