Saudi Arabian Mathematical Competitions

Let
$A=\{x+1, x+2, \ldots, x+7\}$ and $B=\{y+1, y+2, \ldots, y+11\}$, where $x<2004$. We have
$$(x+1)+(x+2)+\ldots+(x+7)=(y+1)+(y+2)+\ldots+(y+11)$$
hence
$$7x+\frac{7 \cdot 8}{2}=11y+\frac{11 \cdot 12}{2}$$
This equation is equivalent to
$$7x-11y=38.$$
The smallest solution in positive integers to this is $(7,1)$ and all solutions are $x=7+11n$, $y=1+7n$, where $n$ is an arbitrary positive integer.
Since $x<2004$, it follows $7+11n<2004$, that is $n<\frac{1997}{11}$. We get $n \leq 181$, hence the maximum possible element of $A$ is $x+7=14+11 \cdot 181=2005$.

Choose
$$A=\{a-3, a-2, a-1, a, a+1, a+2, a+3\}$$
and
$$B=\{b-5, b-4, \ldots, b-1, b, b+1, \ldots, b+4\}$$
We have $7a=11b$, hence $11 \mid a$ and $7 \mid b$. It follows $a=11m$ and $b=7m$, for some positive integer $m$. But $a+3 \leq 2010$ implies $11m \leq 2007$, that is $m \leq \frac{2007}{11}$, hence $m=182$. The desired number is $11 \cdot 182+3=2005$.