2. It is known that the sum of the reciprocals of three consecutive positive integers equals $\frac{191}{504}$. Then the sum of these three numbers is ( ).
(A) 27
(B) 24
(C) 21
(D) 18
2.B.

Let these three consecutive positive integers be $n-1, n, n+1$. Then
$$
\frac{191}{504}=\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}=\frac{3 n^{2}-1}{n\left(n^{2}-1\right)},
$$

which simplifies to $191 n^{3}-1512 n^{2}-191 n+504=0$.
Factoring gives $(n-8)\left(191 n^{2}+16 n-63\right)=0$.
When $n \geqslant 1$, $191 n^{2}+16 n-63>0$. Therefore, $n=8$.