$\frac{2000-x}{4}+\frac{2001-x}{3}=\frac{2-x}{2002}+\frac{1-x}{2003}$
$⇔\frac{2000-x}{4}+1+\frac{2001-x}{3}+1=\frac{2-x}{2002}+1+\frac{1-x}{2003}+1$
$⇔\frac{2004-x}{4}+\frac{2004-x}{3}=\frac{2004-x}{2002}+\frac{2004-x}{2003}$
$⇔(2004-x)(\frac{1}{4}+\frac{1}{3}-\frac{1}{2002}-\frac{1}{2003})=0$
$⇔2004-x=0$
$⇔x=2004$.
