设 $f(k)=m$，则 $|\sqrt[4]{k}-m|<\dfrac{1}{2}$，即 $(m-\dfrac{1}{2})^4<k<(m+\dfrac{1}{2})^4$．而 $(m+\dfrac{1}{2})^4-(m-\dfrac{1}{2})^4=4m^3+m$，因此满足 $f(k)=m$ 的 $k$ 有 $m(4m^2+1)$ 个．注意到 $6^4<2018<7^4$，从而 $f(2018)=6$ 或 $7$．由于 $\displaystyle\sum_{m=1}^{6}m(4m^2+1)=1785$，所以 $f(1786)=f(1787)=\cdots=f(2018)=7$．因此 $\displaystyle\sum_{k=1}^{2018}\dfrac{1}{f(k)}=\displaystyle\sum_{m=1}^{6}(4m^2+1)+\dfrac{2018-1785}{7}=\dfrac{2823}{7}$．