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:
$\{1\}, \quad\{3,5,7\}, \quad\{9,11,13,15,17\}, \cdots$
(First group) (Second group) (Third group)

Then 1991 is in the  group.
3. 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, we get $(n-1)^{2} \leqslant 995$, so $n \leqslant 32$; from the second inequality, we need $2 n^{2} \geqslant 1992$, so $n \geqslant 32$, hence $n=32$.