Problem 4. Find the maximum possible value of the product of different positive integers with sum 2004.
4. Let $x_{1}+x_{2}+\cdots+x_{k}=2004, x_{1}, x_{2}, \ldots, x_{k} \in \mathbb{N}, x_{1}x_{i} x_{j}
$$

a contradiction. Hence $x_{1}, x_{2}, \ldots, x_{k}$ are consecutive integers but at most one.

Let the numbers be $\left\{x_{1}, x_{2}, \ldots, x_{k}\right\}=\{x, x+1, \ldots, x+\ell, x+\ell+n, x+$ $\ell+n+1, \ldots, x+k+n-2\}$ and $k=n+\ell-2$. If $n \geq 3$, we replace $x+\ell$ and $x+\ell+n$ by $x+\ell+1$ and $x+\ell+n-1$, respectively, and, as above, we get a larger product.

Let $n=1$ and the numbers be $x, x+1, \ldots, x+k-1$. If $x \geq 5$ we replace $x$ by the numbers $x-2$ and 2 . The sum remains 2004 and the product increases, since $2(x-2)>x$. If $1 \leq x \leq 4$, a direct verification shows that either we have a larger product or the sum is not equal to 2004 (for $x=2$ and $x=3$ ).

It remains to consider the case $n=2$. Let the numbers be

$$
x, x+1, \ldots, x+\ell, x+\ell+2, x+\ell+3, \ldots, x+k, \quad \ell \geq 0, k \geq \ell+2
$$

As above, we get a larger product for $x=1$ and $x \geq 4$.

If $x=2$, then $2+3+\cdots+(\ell+2)+(\ell+4)+\cdots+(k+2)=2004$ and hence $(k+2)(k+3)=2(2008+\ell)$. Since $0 \leq \ell \leq k-2$, it follows that $4016 \leq(k+2)(k+3) \leq 4012+2 k$ and then $k=61, \ell=8$. So the numbers are $2,3, \ldots, 10,12,13, \ldots, 63$ with product $\frac{63!}{11}$.

For $x=3$ we get analogously $k=60, \ell=5$ and the numbers are $3,4, \ldots, 8,10,11, \ldots, 63$ with product $\frac{63!}{18}$ which is smaller than $\frac{63!}{11}$.