2. Divide the set of positive odd numbers in ascending order into groups, with the $n$-th group containing $n^{2}$ odd numbers. Then 2005 is in the ( )-th group.
(A) 14
(B) 15
(C) 16
(D) 17
2. A.

Consider the sequence of groups formed by adding 1 to each odd number and then dividing by 2:
$$
\{1\},\{2,3,4,5\}, \cdots \text {. }
$$

Thus, 2005 corresponds to 1003 at this point, so we only need to consider the group where 1003 is located.
Assume 1003 is in the $n$-th group, then we have
$$
\begin{array}{l}
1^{2}+2^{2}+\cdots+(n-1)^{2}<1003 \\
\leqslant 1^{2}+2^{2}+\cdots+n^{2},
\end{array}
$$

which means $(n-1) n(2 n-1)<6018 \leqslant n(n+1)(2 n+1)$. From this, we know $n=14$, meaning 2005 is in the 14th group.