6. In the sequence of positive integers, starting from 1, certain numbers are painted red according to the following rules. First, paint 1, then paint 2 even numbers $2, 4$; then paint the 3 nearest consecutive odd numbers after 4, which are $5, 7, 9$; then paint the 4 nearest consecutive even numbers after 9, which are $10, 12, 14, 16$; then paint the 5 nearest consecutive odd numbers after 16, which are $17, 19, 21, 23, 25$. Continue this process to get a red subsequence $1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17, \cdots$. In this red subsequence, the 2003rd number starting from 1 is ( ).
(A) 3844
(B) 3943
(C) 3945
(D) 4006
6. (B).

The first time, 1 number is painted red: $1,1=1^{2}$;
The second time, 2 numbers are painted red: $2,4,4=2^{2}$;
The third time, 3 numbers are painted red: $5,7,9,9=3^{2}$.
It is easy to see that the last number painted red in the $k$-th time is $k^{2}$, then the $k+1$ numbers painted red in the $k+1$-th time are:
$$
k^{2}+1, k^{2}+3, k^{2}+5, \cdots, k^{2}+2 k-1, k^{2}+2 k+1 \text {, }
$$

where the last number is $(k+1)^{2}$. According to this rule, the first $k$ segments have painted a total of
$$
1+2+3+\cdots+(k-1)+k=\frac{k(k+1)}{2}
$$

numbers.
Solving the inequality $\frac{k(k+1)}{2} \leqslant 2003$, we get $k \leqslant 62$.
At this point, the first 62 segments have a total of $\frac{62 \times 63}{2}=1953$ numbers. The last number in the 62nd segment is $62^{2}=3844$, which is the 1953rd red number. The 2003rd red number should be in the 63rd segment, and all numbers in the 63rd segment are odd.
Since $2003-1953=50$, the 2003rd red number is
$$
3844+1+(50-1) \times 2=3943 .
$$