12. B. Let the positive integer $a$ satisfy $192 \mid\left(a^{3}+191\right)$, and $a<2$ 009. Find the sum of all possible positive integers $a$ that meet the condition.
12. B. From the problem, we have $192 \mid\left(a^{3}-1\right)$.
$$
\begin{array}{l}
\text { Also, } 192=3 \times 2^{6}, \text { and } \\
a^{3}-1=(a-1)[a(a+1)+1] \\
=(a-1) a(a+1)+(a-1) .
\end{array}
$$

Since $a(a+1)+1$ is odd, we have
$$
2^{6}\left|\left(a^{3}-1\right) \Leftrightarrow 2^{6}\right|(a-1) \text {. }
$$

Also, $3 \mid(a-1) a(a+1)$, so
$$
3\left|\left(a^{3}-1\right) \Leftrightarrow 3\right|(a-1) \text {. }
$$

Therefore, $192 \mid(a-1)$. Thus, $a=192 k+1$.
Given $0<a<2009$, then $k=0,1, \cdots, 10$.
Hence, the sum of all possible positive integers $a$ that satisfy the condition is $11+192(1+2+\cdots+10)=10571$.