$\frac{x+4}{2000} + \frac{x+2}{2002} > \frac{x+83}{1921} + \frac{x+104}{1900}$
$\Leftrightarrow \left(\frac{x+4}{2000} + 1\right) + \left(\frac{x+2}{2002} + 1\right) > \left(\frac{x+83}{1921} + 1\right) + \left(\frac{x+104}{1900} + 1\right)$
$\Leftrightarrow \frac{x+4+2000}{2000} + \frac{x+2+2002}{2002} > \frac{x+83+1921}{1921} + \frac{x+104+1900}{1900}$
$\Leftrightarrow \frac{x+2004}{2000} + \frac{x+2004}{2002} > \frac{x+2004}{1921} + \frac{x+2004}{1900}$
$\Leftrightarrow \frac{x+2004}{2000} + \frac{x+2004}{2002} - \frac{x+2004}{1921} - \frac{x+2004}{1900} > 0$
$\Leftrightarrow (x+2004)\left(\frac{1}{2000} + \frac{1}{2002} - \frac{1}{1921} - \frac{1}{1900}\right) > 0$
Đặt $A = \frac{1}{2000} + \frac{1}{2002} - \frac{1}{1921} - \frac{1}{1900}$
Ta có $\frac{1}{2000} < \frac{1}{1921}$ và $\frac{1}{2002} < \frac{1}{1900}$
Do đó $\frac{1}{2000} + \frac{1}{2002} < \frac{1}{1921} + \frac{1}{1900}$
$\Leftrightarrow \frac{1}{2000} + \frac{1}{2002} - \frac{1}{1921} - \frac{1}{1900} < 0$
Vậy $A < 0$.
Để $(x+2004)A > 0$ mà $A < 0$, ta phải có $x+2004 < 0$.
$\Leftrightarrow x < -2004$
Vậy tập nghiệm của bất phương trình là $S = (-\infty; -2004)$.