9. The set of positive odd numbers $\{1,3,5, \cdots\}$ is grouped in ascending order such that the $n$-th group contains $(2 n-1)$ odd numbers:
$$
\begin{array}{l}
\{1\}, \quad\{3,5,7\}, \quad\{9,11,13,15,17\}, \cdots \\
\text { (first group) (second group) (third group) } \\
\end{array}
$$

Then 1991 is in the $\qquad$ group.
$1991=2 \cdot 996-1$ is the 996th odd number, and $1+3+5+\cdots+(2 n-1)=n^{2}$. Since $31^{2}<996<32^{2}$, 1991 is located in the 32nd group.