Saudi Arabia Mathematical Competitions

Let $A=\{x+1,x+2,\dots ,x+7\}$ and $B=\{y+1,y+2,\dots ,y+11\}$, where $x<2004$. We have $$(x+1)+(x+2)+\dots +(x+7)=(y+1)+(y+2)+\dots +(y+11)$$ 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 $n\le 181$, hence the maximum possible element of $A$ is $x+7=14+11\cdot 181=2005$.

---

Alternative solution.

Choose $$A=\{a-3,a-2,a-1,a,a+1,a+2,a+3\}$$ and $$B=\{b-5,b-4,\dots ,b-1,b,b+1,\dots ,b+4\}$$ 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 $a+3\le 2010$ implies $11m\le 2007$, that is $m\le \frac{2007}{11}$, hence $m=182$. The desired number is $11\cdot 182+3=2005$.