1. The sequence of positive integers $1, 2, \cdots$ is divided from left to right into segments such that the first segment has $1 \times 2$ numbers, the second segment has $2 \times 3$ numbers, $\cdots$, the $n$-th segment has $n \times (n+1)$ numbers, $\cdots$. Then 2014 is in the $\qquad$ segment.
$-, 1.18$
$$
\begin{array}{l}
\text { Let } S_{n}=1 \times 2+2 \times 3+\cdots+n(n+1) \\
=\left(1^{2}+2^{2}+\cdots+n^{2}\right)+(1+2+\cdots+n) \\
=\frac{n(n+1)(n+2)}{3} .
\end{array}
$$

If 2014 is in the $(n+1)$-th segment, since there are $S_{n}$ numbers before this segment, then $S_{n}<2014 \leqslant S_{n+1}$.
Since $S_{17}=1938, S_{18}=2280$, and
$$
S_{17}<2014 \leqslant S_{18} \text {, }
$$

Therefore, 2014 is in the 18th segment.