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

Then the number in 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.D.

From the observation, we can deduce the arrangement characteristics of this natural number table:
(1) Every number in the 1st column is a perfect square, and it is exactly equal to the square of its row number, i.e., the 1st number in the $n$-th row is $n^{2}$;
(2) The $n$-th number in the 1st 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 at the 2002nd row from the top and the 2003rd column from the left should be the 2002nd number in the 2003rd column, which is
$$
\begin{array}{l}
{\left[(2003-1)^{2}+1\right]+2001} \\
=2002^{2}+2002=2002 \times 2003 .
\end{array}
$$