Which are the two different natural numbers, at least two digits each, for which the sum of their sum, positive difference, product, and quotient is 576.
According to the problem, the quotient of the two numbers, when added to integers, results in an integer, so the quotient must be an integer. Therefore, the two numbers are in the form of $x$ and $xy$, where $x$ and $y$ are integers.

According to the problem,

$$
(xy+x)+(xy-x)+x^{2}y+y=576
$$

which means

$$
2xy+x^{2}y+y=y(x^{2}+2x+1)=y(x+1)^{2}=576
$$

From this,

$$
y=\frac{576}{(x+1)^{2}}=\left(\frac{24}{x+1}\right)^{2}
$$

The expression in the parentheses on the right side must be an integer, otherwise its square would not be an integer. Thus, $x+1$ must be a divisor of 24, and furthermore, it must be a two-digit number because $x$ itself is two-digit, so $x+1$ is either 12 or 24. In the latter case, $x=23$ and $y=1$, and thus $xy$ is also 23, which is excluded by the problem statement.

In the former case,

$$
x=11, \quad y=\left(\frac{24}{12}\right)^{2}=4, \quad \text{and thus} \quad xy=44
$$

These two different two-digit numbers meet the requirements of the problem.

Medek Béla (Bp. II., Rákóczi g. II. o. t.)