Let $f(x)=|2x-1|$ where $x$ denotes the fractional part of $x$. The number $n$ is the smallest positive integer such that the equation $nf(xf(x))=x$ has at least $3000$ real solutions. What is $n$?

Note: the fractional part of $x$ is a real number $y=x$ such that $0\le y<1$ and $x-y$ is an integer.