Answer: 400

Small Hint:

Count the integers $n$ for which $j-\tfrac12<\sqrt[4]n<j+\tfrac12$

Big Hint:

The number of occurrences of $f(n)=j$ simplifies to $4j^3+j$

Solution:

For $j\geq1,$ the number of positive integers $n$ for which $f(n)=j$ is $$\begin{aligned}\left(j+\frac12\right)^4&-\left(j-\frac12\right)^4\\&=4j^3+j.\end{aligned}$$ For $j=1,\ldots,6,$ these account for $$\sum_{j=1}^6(4j^3+j)=1785$$ values, and their contribution to the requested sum is $$\begin{aligned}\sum_{j=1}^6\frac{4j^3+j}{j}&=4\sum_{j=1}^6j^2+6\\&=370.\end{aligned}$$ The remaining $1995-1785=210$ values have $f(n)=7,$ contributing $30.$ The total is $400.$