6. As shown in Figure 2,

A particle moves in the first quadrant. In the first second, it moves from the origin to $(0,1)$, and then it continues to move back and forth in directions parallel to the $x$-axis and $y$-axis, moving one unit length each time. Therefore, after 1999 seconds, the position of the particle is ( ).
(A) $(44,25)$
(B) $(45,26)$
(C) $(26,45)$
(D) $(25,44)$
6. (D).

Since $1+3+5+\cdots+(2 n-1)=n^{2}$.
When $n$ is odd, the particle's position after $n^{2}$ seconds is $(0, n)$, moving to the right.

When $n$ is even, the particle's position after $n^{2}$ seconds is $(n, 0)$, moving to.
$$
\because 1+3+\cdots+(2 \times 44-1)=c, 4^{2}=1936 \text {, }
$$
$\therefore 1936$ seconds later, the particle's position is approximately $(14,0)$, direction upwards. Therefore, the particle's position after 1999 seconds should be
$$
(44-(1999-1936-44), 44) .
$$