Yanıt: $\boxed{C}$
$\begin{array}{lcl}
\sum\limits_{k=1}^{n} k \cdot (n+1-k) &=& (n+1)\sum\limits_{k=1}^{n} k - \sum\limits_{k=1}^{n} k^2 \\
&=& \dfrac{(n+1) \cdot n \cdot (n+1)}{2} - \dfrac{n \cdot (n+1) \cdot (2n+1)}{6} \\
&=& \dfrac{n(n+1)}{6} \cdot (3n+3 - 2n -1) \\
&=& \dfrac{n(n+1)(n+2)}{6} \\
\end{array}$
$n=2003$ için
$\sum\limits_{k=1}^{2003} k \cdot (2004-k) = \dfrac{2003 \cdot 2004 \cdot 2005}{6} = 334 \cdot 2003 \cdot 2005 = 2 \cdot 5 \cdot 167 \cdot 401 \cdot 2003$.