2. The number of integers $x$ that satisfy the inequality $|x-2006|+|x| \leqslant 9999$ is ( ) .
(A) 9998
(B) 9999
(C) 10000
(D) 10001
2.B.

When $x \geqslant 2006$, we have $x-2006+x \leqslant 9999$, which means $x \leqslant \frac{2006+9999}{2}$, or $x \leqslant 6002$;

When $0 \leqslant x<2006$, we have $2006-x+x \leqslant 9999$, which is obviously true;

When $x<0$, we have $2006-x-x \leqslant 9999$, which means $x \geqslant \frac{2006-9999}{2}$, or $x \geqslant -3996$.
In summary, $-3996 \leqslant x \leqslant 6002$.
Therefore, there are 9999 integers $x$ that satisfy the inequality.