2. Find all pairs of natural numbers $m$ and $n$ whose sum is 2007, and whose product is divisible by 2007.
We can write the equations $m+n=2007$ and $m n=2007 a$, where $a$ is some natural number. Then $n=2007-m$ and we get $m(2007-m)=2007 a$ or $2007(m-a)=m^{2}$. Therefore, the number $m^{2}$ is divisible by the number $2007=3^{2} \cdot 223$. From this, it follows that $m$ is divisible by $3 \cdot 223$. Let's write $m=3 \cdot 223 \cdot k$, where $k$ is some natural number. Since $m<2007=3 \cdot 3 \cdot 223$, it follows that $k<3$. Therefore, either $m=3 \cdot 223 \cdot 1=669$ or $m=3 \cdot 223 \cdot 2=1338$. In the first case, $n=1338$, and in the second case, $n=669$.

![](https://cdn.mathpix.com/cropped/2024_06_07_f155edc146c61fd862beg-06.jpg?height=303&width=1639&top_left_y=2207&top_left_x=217)