$\begin{array}{l} \frac{{x + 4}}{{2000}} + \frac{{x + 3}}{{2001}} = \frac{{x + 2}}{{2002}} + \frac{{x + 1}}{{2003}}\\ = > \frac{{x + 4}}{{2000}} + 1 + \frac{{x + 3}}{{2001}} + 1 = \frac{{x + 2}}{{2002}} + 1 + \frac{{x + 1}}{{2003}} + 1\\ = > \frac{{x + 2004}}{{2000}} + \frac{{x + 2004}}{{2001}} = \frac{{x + 2004}}{{2002}} + \frac{{x + 2004}}{{2003}}\\ = > \frac{{x + 2004}}{{2000}} + \frac{{x + 2004}}{{2001}} - \frac{{x + 2004}}{{2002}} - \frac{{x + 2004}}{{2003}} = 0\\ = > \left( {x + 2004} \right)\left( {\frac{1}{{2002}} + \frac{1}{{2001}} - \frac{1}{{2002}} - \frac{1}{{2003}}} \right) = 0\\ = > x + 2004 = 0\left( {do:\frac{1}{{2002}} + \frac{1}{{2001}} - \frac{1}{{2002}} - \frac{1}{{2003}} \ne 0} \right)\\ = > x = - 2004\\ vay:x = - 2004 \end{array}$