8. Given positive integers $a_{1}, a_{2}, \cdots, a_{2018}$ satisfying
$$
a_{1}<a_{2}<\cdots<a_{2018} \text {. }
$$

For $i=1,2, \cdots, 2018, b_{i}$ is the number of positive integers in $a_{1}, a_{2}, \cdots, a_{2018}$ that do not exceed $i$. Then
$$
\frac{\sum_{k=1}^{2018} a_{k}+\sum_{k=1}^{a_{2018}} b_{k}}{a_{2018}+1}=
$$
$\qquad$
8. 2018.

Notice that, when $k < a_{1}$, $b_{k} = 0$;
when $a_{j} \leqslant k < a_{j+1} (j=1,2, \cdots, 2017)$, $b_{k} = j$;
when $k = a_{2018}$, $b_{k} = 2018$.
$$
\begin{array}{l}
\text{Thus } \sum_{k=1}^{a_{2018}} b_{k} = \sum_{j=1}^{2017} j\left(a_{j+1} - a_{j}\right) + 2018 \\
= -\sum_{k=1}^{2018} a_{k} + 2018\left(a_{2018} + 1\right).
\end{array}
$$

Therefore, $\frac{\sum_{k=1}^{2018} a_{k} + \sum_{k=1}^{a_{2018}} b_{k}}{a_{2018} + 1} = 2018$.