10. Let $f(n)$ be the integer closest to $\sqrt[4]{n}$. Then $\sum_{k=1}^{2018} \frac{1}{f(k)}=$ $\qquad$
10. $\frac{2823}{7}$.

Let $f(k)=m$. Then
$$
|\sqrt[4]{k}-m|<\frac{1}{2},
$$

which means $\left(m-\frac{1}{2}\right)^{4}<k<\left(m+\frac{1}{2}\right)^{4}$.
And $\left(m+\frac{1}{2}\right)^{4}-\left(m-\frac{1}{2}\right)^{4}=4 m^{3}+m$, so there are $m\left(4 m^{2}+1\right)$ values of $k$ that satisfy $f(k)=m$.
Notice that, $6^{4}<2018<7^{4}$.
Thus, $f(2018)=6$ or 7.
Since $\sum_{m=1}^{6} m\left(4 m^{2}+1\right)=1785$, we have,
$$
\begin{array}{l}
f(1786)=f(1787)=\cdots=f(2018)=7 . \\
\text { Therefore, } \sum_{k=1}^{2018} \frac{1}{f(k)} \\
=\sum_{m=1}^{6}\left(4 m^{2}+1\right)+\frac{2018-1785}{7} \\
=\frac{2823}{7} .
\end{array}
$$