jmc

We have $f(n)=m$ if and only if $$m-\frac{1}{2}<\sqrt[4]{n}<m+\frac{1}{2},$$or $${\left(m-\frac{1}{2}\right)}^4<n<{\left(m+\frac{1}{2}\right)}^4.$$Expanding the fourth powers, we get $$m^4-2m^3+\frac{3}{2}{m}^2-\frac{1}{2}m+\frac{1}{16}<n<m^4+2m^3+\frac{3}{2}{m}^2+\frac{1}{2}m+\frac{1}{16}.$$The leftmost and rightmost expressions are both non-integers, and their difference is $4m^3+m$. Therefore, there are exactly $4m^3+m$ values of $n$ that satisfy this inequality.

For each $m$, there are $4m^3+m$ terms of the form $\frac{1}{m}$ in the sum, so those terms contribute $(4m^3+m)\cdot \frac{1}{m}=4m^2+1$ to the sum. Thus, from $m=1$ to $m=6$, we get $4(1+4+9+16+25+36)+6=370$.

The remaining terms have $m=7$. Since $6.5^4=1785\frac{1}{16}$, these are the terms from $n=1786$ to $n=1995$, inclusive. There are $1995-1786+1=210$ such terms, so they contribute $210\cdot \frac{1}{7}=30$ to the sum. Therefore, the final answer is $370+30=\boxed{400}$.