1. Divide the set of positive even numbers $\{2,4, \cdots\}$ into groups in ascending order, with the $n$-th group containing $3 n-2$ numbers:
$$
\{2\},\{4,6,8,10\},\{12,14, \cdots, 24\}, \cdots \text {. }
$$

Then 2018 is in the group.
- 1. 27 .

Let 2018 be in the $n$-th group. Since 2018 is the 1009th positive even number and according to the problem, we have
$$
\begin{array}{l}
\sum_{i=1}^{n-1}(3 i-2)<1009 \leqslant \sum_{i=1}^{n}(3 i-2) \\
\Rightarrow \frac{3(n-1)^{2}-(n-1)}{2}<1009 \leqslant \frac{3 n^{2}-n}{2} \\
\Rightarrow n=27 .
\end{array}
$$

Therefore, 2018 is located in the 27th group.