Let's find the positive integers $A, B$ such that writing their decimal representations consecutively forms a perfect square, and this is equal to twice the product of $A$ and $B$.
A square number can only end in an even number of zeros. Let's write the number $B$ after the number $A$, so $B$ must end in an even number of zeros since $\overline{A B}$ is a square number. It is easy to see that if a pair $A, B$ satisfies the problem's requirements, then $A, B \cdot 100$ also satisfies them. (Since both $\overline{A B}$ and $2 \cdot A \cdot B$ become 100 times larger.) Therefore, we can assume that $B$ is not divisible by 10. Let the number $B$ be $b$-digit. Then $\overline{A B}$ is a square number, and

$$
\overline{A B}=10^{b} \cdot A+B=2 \cdot A \cdot B
$$

Rearranging,

$$
10^{b} \cdot A=(2 A-1) \cdot B
$$

follows, which means $2^{b} \cdot A$ is a divisor of the product $(2 A-1) \cdot B$. We claim that $2^{b} \cdot A$ and the first factor of the product, $2 A-1$, are coprime. Indeed, any prime divisor $p$ of $2^{b} \cdot A$ is either $p=2$ or $p$ is a divisor of $A$ and thus of $2 A$ as well. The number 2 is clearly not a divisor of the odd number $(2 A-1)$, and in the second case, $p$ cannot be a divisor of both the adjacent numbers $(2 A-1)$ and $2 A$.

Thus, since $2^{b} \cdot A$ is a divisor of the product $(2 A-1) \cdot B$, it is also a divisor of the second factor, $B$, i.e., $B=k \cdot 2^{b} \cdot A$ for some positive integer $k$. Simplifying (1) by $2^{b} \cdot A$,

$$
5^{b}=(2 A-1) \cdot k
$$

follows. From (2), both $(2 A-1)$ and $k$ are powers of 5. Since $b \geq 1$, the value of $k$ can only be 1; otherwise, $B=k \cdot 2^{b} A$ would be divisible by 10.

Thus, $5^{b}=2 A-1$, so $A=\frac{5^{b}+1}{2}$ and $B=2^{b} \cdot A$. Then $2 \cdot A \cdot B=2^{b+1} A^{2}$ is a square number if and only if $b$ is odd. Therefore, $A=\frac{5^{b}+1}{2}$ and $B=2^{b} \cdot A=2^{b-1}\left(5^{b}+1\right)$, where $b$ is odd.

Thus, $2 \cdot A \cdot B=2^{b+1} A^{2}$ is indeed a square number, and the equation $\overline{A B}=2 \cdot A \cdot B$ holds, since

$$
10^{b} A+B=10^{b} \cdot A+2^{b} \cdot A=2^{b} \cdot A\left(5^{b}+1\right)=2^{b+1} \cdot A^{2}=2 \cdot A \cdot B
$$

Finally, $10^{b-1}<B<10^{b}$, so $B$ indeed has $b$ digits.

Therefore, the numbers of the form

$$
A=\frac{5^{b}+1}{2} \quad \text{and} \quad B=2^{b} \cdot A \cdot 100^{m}
$$

where $b$ is odd and $m$ is any natural number, satisfy the problem's requirements.