jmc

The graph of $y=f(x)$ is shown below.



In particular, $0\le f(x)\le 1$ for all $x.$ So, $$0\le nf(xf(x))\le n,$$which means that all solutions to $nf(xf(x))=x$ lie in the interval $[0,n].$

Let $a$ be an integer such that $0\le a\le n-1.$ Suppose $a\le x<a+\frac{1}{2}.$ Then $$f(x)=|2\{x\}-1|=|2(x-a)-1|=1+2a-2x.$$Let $$g(x)=xf(x)=x(1+2a-2x).$$Thus, we want to find the solutions to $f(g(x))=\frac{x}{n}.$

If $a=0,$ then $$g(x)=x(1-2x),$$which satisfies $0\le g(x)\le \frac{1}{8}$ for $0\le x<\frac{1}{2}.$ Then $$f(g(x))=1-2g(x)=4x^2-2x+1.$$We can check that $$\frac{3}{4}\le 4x^2-2x+1\le 1$$for $0\le x<\frac{1}{2}.$ But $\frac{x}{n}\le \frac{1}{2},$ so there no solutions in this case.

Otherwise, $a\ge 1.$ Suppose $a\le x<y<a+\frac{1}{2}.$ We claim that $g(x)>g(y).$ This inequality is equivalent to $$x(1+2a-2x)>y(1+2a-2y),$$which in turn is equivalent to $(y-x)(2x+2y-2a-1)>0.$ Since $2x+2y-2a-1>2a-1\ge 1,$ the claim $g(x)>g(y)$ is established.

This means that $g(x)$ is strictly decreasing on the interval $a\le x<a+\frac{1}{2},$ so it maps the interval $\left[a,a+\frac{1}{2}\right)$ bijectively to the interval $(0,a].$ This means that $f(g(x))$ oscillates between 0 and 1 $2a$ times, so the line $y=\frac{x}{n}$ intersects this graph $2a$ times.

Now suppose $a+\frac{1}{2}\le x<a.$ Then $$f(x)=|2\{x\}-1|=|2(x-a)-1|=2x-2a-1.$$Let $$g(x)=xf(x)=x(2x-2a-1).$$We can similarly establish that $g(x)$ is strictly increasing for $a+\frac{1}{2}\le x<a,$ so it maps the interval $\left[a+\frac{1}{2},a\right)$ bijectively to the interval $[0,a+1).$ This means that $f(g(x))$ oscillates between 0 and 1 $2a+2$ times, so the line $y=\frac{x}{n}$ intersects this graph $2a+2$ times.

Therefore, the total number of solutions is $$\sum _{a=0}^{n-1}(2a+2a+2)=2\sum _{a=0}^{n-1}(2a+1)=2n^2.$$Finally, the smallest such $n$ such that $2n^2\ge 2012$ is $n=\boxed{32}.$