Bulgarian Mathematical Competitions

Solution:
Let $x_{1} + x_{2} + \cdots + x_{k} = 2004$, $x_{1}, x_{2}, \ldots, x_{k} \in \mathbb{N}$, $x_{1} < x_{2} < \cdots < x_{k}$ and the product $x_{1} x_{2} \ldots x_{k}$ is maximal. Assume that for some $i, j$, $1 \leq i < j \leq k$ one has that $x_{i} \leq x_{i+1} - 2$ and $x_{j} \leq x_{j+1} - 2$. Then replacing $x_{i}$ and $x_{j}$ by $x_{i} + 1$ and $x_{j} - 1$, respectively (the sum is the same, i.e. $2004$), we get a larger product since
$$(x_{i} + 1)(x_{j} - 1) = x_{i} x_{j} + x_{j} - x_{i} - 1 > x_{i} x_{j}$$
a contradiction. Hence $x_{1}, x_{2}, \ldots, x_{k}$ are consecutive integers but at most one.

Let the numbers be $\{x_{1}, x_{2}, \ldots, x_{k}\} = \{x, x+1, \ldots, x+\ell, x+\ell+n, x+\ell+n+1, \ldots, x+k+n-2\}$ 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 $4016 \leq (k+2)(k+3) \leq 4012+2k$ 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 $\frac{63!}{11}$.