3. 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: (first group) $\{1\}$, (second group) $\{3,5,7\}$, (third group) $\{9,11,13,15,17\}, \cdots$ Then 1991 is in the  group.
32
3.【Analysis and Solution】Since $1+3+5+\cdots+(2 n-1)=n^{2}$, the last number of the $n$-th group, which is the $n^{2}$-th odd number, is $2 n^{2}-1$. Therefore, we have the inequality $2(n-1)^{2}+1 \leqslant 1991 \leqslant 2 n^{2}-1$.
From the first inequality, $(n-1)^{2} \leqslant 995$, so $n \leqslant 32$; from the second inequality, $2 n^{2} \geqslant 1992$, so $n \geqslant 32$, hence $n=$
32.