答 27 .
Analysis: Suppose 2018 is in the $n$-th group. Since 2018 is the 1009th positive even number, and according to the problem, we have
$$
\sum_{i=1}^{n-1}(3 i-2)<1009 \leq \sum_{i=1}^{n}(3 i-2)
$$

That is,
$$
\frac{3(n-1)^{2}-(n-1)}{2}<1009 \leq \frac{3 n^{2}-n}{2}
$$

Solving for the positive integer $n=27$, so 2018 is in the 27th group.
1. Divide the set of positive even numbers $\{2,4,6, \cdots\}$ into groups in ascending order, with the $n$-th group containing $3 n-2$ numbers:
$$
\{2\},\{4,6,8,10\}, \quad\{12,14,16,18,20,22,24\}, \cdots
$$

Then 2018 is in the $\qquad$ group.