Is 2004 a positive integer such that the sum of some positive integers is equal to their product?
Solution. If each of the 2004 numbers were 1, then the sum of the numbers would be 2004, and their product would be 1, which does not meet the requirements of the problem. Therefore, there must be at least one number different from 1 among the numbers. If there is only one such number, denoted by $x$, then according to the condition, $2003 + x = x$. Thus, the problem has no solution in this case.

Suppose there are 2 numbers different from 1 among the numbers, denoted by $x$ and $y$. Then,

$$
2002 + x + y = xy
$$

Rearranging the equation:

$$
2002 + y = x(y - 1), \quad \text{hence} \quad x = \frac{2002 + y}{y - 1} = \frac{2003 + y - 1}{y - 1} = 1 + \frac{2003}{y - 1}
$$

This is an integer if $y \neq 1$ and $\frac{2003}{y - 1}$ is an integer. Since 2003 is a prime number, two cases are possible:

$\frac{2003}{y - 1} = 1$, hence $2003 = y - 1$, so $y = 2004$ and $x = 2$. Substituting into (1): $2002 + 2004 + 2 = 2 \cdot 2004$, the equality indeed holds.

Or: $\frac{2003}{y - 1} = 2003$, hence $y = 2$ and $x = 2004$. We have already shown that this is a solution to equation (1).

Thus, the answer to the question posed in the problem is: yes. One solution is: 2002 ones, one 2, and one 2004.

Remark: Péter Hutyán (Győr, Czuczor Gergely Bencés Gimn., 11th grade) also examined whether there are other solutions to the problem. In the case of three numbers different from 1 (similar to the previous case), he found that the numbers

$$
\underbrace{1,1,1, \ldots, 1}_{2001 \text{db}}, 10, 12, 17
$$

are also valid. Furthermore, for four numbers different from 1, the numbers

$\underbrace{1,1,1, \ldots, 1}_{2000 \text{db}}, 3, 2, 9, 38$

also satisfy the requirements of the problem. He used a program to solve the corresponding Diophantine equations.