6. Natural numbers are arranged according to the following pattern:
\begin{tabular}{cccc}
$1-2$ & 5 & 10 & 17 \\
$\mid$ & $\mid$ & $\mid$ & $\mid$ \\
$4-3$ & 6 & 11 & 18 \\
& $\mid$ & $\mid$ & $\mid$ \\
$9-8-7$ & 12 & 19 \\
& $\mid$ & $\mid$ \\
$16-15-14-13$ & 20 \\
$25-24-23-22-21$
\end{tabular}

Then the number at the 2002nd row from the top and the 2003rd column from the left is ( ).
A. $2002^{2}$
B. $2003^{2}$
C. $2002+2003$
D. $2002 \times 2003$
6. By observation, we can determine the arrangement characteristics of this natural number table: (1) Every number in the first column is a perfect square, and it is exactly equal to the square of its row number, i.e., the first number in the $n$-th row is $n^{2}$; (2) The $n$-th number in the first row is $(n-1)^{2}+1$; (3) In the $n$-th row, the numbers from the 1st to the $n$-th number decrease by 1 sequentially; (4) In the $n$-th column, the numbers from the 1st to the $n$-th number increase by 1 sequentially. Therefore, the number in the 2002nd row and 2003rd column, starting from the top, is the 2002nd number in the 2003rd column, which is $\left[(2003-1)^{2}+1\right]+2001=2002^{2}+2002=2002 \times 2003$. Hence, the answer is D.