1. Given the sequence $\left\{a_{n}\right\}$ satisfies for any positive integer $n$, $\sum_{k=1}^{n} a_{k}=n^{3}$. Then $\sum_{k=2}^{2017} \frac{1}{a_{k}-1}=$ $\qquad$
$$
-1 \cdot \frac{672}{2017} \text {. }
$$

When $n \geqslant 2$, we have
$$
\sum_{k=1}^{n} a_{k}=n^{3}, \sum_{k=1}^{n-1} a_{k}=(n-1)^{3} \text {. }
$$

Then $a_{n}=3 n^{2}-3 n+1$
$$
\begin{array}{l}
\Rightarrow \frac{1}{a_{n}-1}=\frac{1}{3 n(n-1)} \\
\quad=\frac{1}{3}\left(\frac{1}{n-1}-\frac{1}{n}\right)(n=2,3, \cdots) . \\
\text { Therefore } \sum_{k=2}^{2017} \frac{1}{a_{k}-1}=\frac{1}{3}\left(1-\frac{1}{2017}\right)=\frac{672}{2017} .
\end{array}
$$