VII-4. The difference between a certain two-digit natural number and the number written with the same digits but in reverse order is a square of a natural number or zero. Determine all two-digit natural numbers that have this property.
Solution: According to the condition of the problem, we have:

$\overline{a b} \cdot \overline{b a}=k^{2}$, i.e., $10 a+b-(10 b-a)=k^{2}, a, b \in\{1,2, \ldots, 9\}, 9(a-b)=k^{2}$.

Since $9=3^{2}$, it follows that $a-b$ is a perfect square of some number.

Thus, $a-b=0, a-b=1, a-b=4$ or $a-b=9$.

$1^{\circ}$ If $a-b=0$, i.e., $a=b$, the numbers are: $11,22,33,44,55,66,77,88,99$.

$2^{\circ}$ If $a-b=1$, i.e., $a=b+1$. the numbers are: $10,21,32,43,54,65,76,87,98$.

$3^{\circ}$ If $a-b=4$, i.e., $a=b+4$, the numbers are: $40,51,62,73,84,95$.

$4^{\circ}$ If $a-b=9$, i.e., $a=b+9$, the number is: 90. Thus, in total there are $9+9+6+1=25$ numbers.