If $1^2+2^2+3^2+\cdots +{2003}^2=\left(2003\right)\left(4007\right)$ and (1) (2003) + (2) (2002) + (3) (2001) + ... + (2003) (1) = (2003) (334) (x), then x equals

$\begin{array}{c}\mathrm{S}=\left(1\right)\left(2003\right)+\left(2\right)\left(2002\right)+\left(3\right)\left(2001\right)+\cdots +\left(2003\right)\left(1\right)\\ =\sum _{\mathrm{r}=1}^{2003} \mathrm{r}(2003-(\mathrm{r}-1\left)\right)\\ =\sum _{\mathrm{r}=1}^{2003} \mathrm{r}(2004-\mathrm{r})\\ =\sum _{\mathrm{r}=1}^{2003} 2004\mathrm{r}-\sum _{\mathrm{r}=1}^{2003} {\mathrm{r}}^2\\ =\frac{2004\times 2003\times 2004}{2}-2003\times 4007\times 334\\ =2003\times 334\times (6012-4007)\\ =2003\times 334\times 2005\end{array}$

Hence, x = 2005.