75. There is a wonderful positive integer $A$, which is composed of only two different digits. However, $A^{2}$ contains exactly the 10 digits 0, 1, 2, ..., 9. Try to find the positive integer $A$.
Since $A^{2}$ contains exactly the digits $0,1,2, \cdots, 9$, we have
$$
\begin{array}{l}
1023456789 \leqslant A^{2} \leqslant 9876543210 . \\
\therefore 31992 \leqslant A \leqslant 99380 .
\end{array}
$$

Therefore, $A$ is a five-digit number, and the unit digit of $A$ is not 0.
$$
\text { Also, } \because A^{2} \equiv 0+1+2+\cdots+9 \equiv 45 \equiv 0(\bmod 9) \text {, }
$$
$\therefore A^{2}$ is a multiple of 9. Hence, $A$ is a multiple of 3.
Assume $A$ is composed of two different digits $a, b(0 \leqslant a, b \leqslant 9)$.
(1) If $A$ contains one digit $a$ and four digits $b$, then
$$
A \equiv a+4 b \equiv a+b \equiv 0(\bmod 3) \text {. }
$$

If $a=0$, then $b=3,6,9$.
Thus, the possible values of $A$ are $33303,33033,30333$, $66606,66066,60666,99909,99099,90999$.
Upon verification, none of these 9 numbers satisfy the condition.
Similarly, it can be proven that when $a$ is any other value, there are no numbers that satisfy the condition.

Therefore, when $A$ contains one digit $a$ and four digits $b$, there are no numbers that satisfy the condition.
(2) When $A$ contains two digits $a$ and three digits $b$, then
$$
A \equiv 2 a+3 b \equiv 2 a \equiv 0(\bmod 3) \text {. }
$$

Thus, $a$ is a multiple of 3.
When $a=0$, $b=1,2, \cdots, 9$.
The possible values of $A$ are $11001,10101,10011,22002$, $20202,20022, \cdots, 90099$.
Upon verification, none of these numbers satisfy the condition.
Similarly, when $a=3$, there are no numbers that satisfy the condition.
When $a=6$, there is one number that satisfies the condition: 67677,
$$
67677^{2}=4580176329 \text {; }
$$

When $a=9$, there is one number that satisfies the condition: 97779, $97779^{2}=9560732841$.

In summary, the required positive integers are 67677, 97779.
(Zhang Yanwei, Suqian City Education Commission, Jiangsu Province, 223800)