Solution:

Since $0 \le f(x) \le 1,$ every solution lies in $[0, n].$ The function $f$ is a triangular wave of period $1,$ and $g(x) = xf(x)$ is monotonic on each half-integer interval, mapping it onto an interval on which $f(g(x))$ oscillates.

Counting the oscillations, on the intervals $[a, a + \tfrac12)$ and $[a + \tfrac12, a+1)$ the curve $y = f(g(x))$ meets the line $y = \tfrac{x}{n}$ a total of $2a$ and $2(a+1)$ times. Summing over $a = 0, \ldots, n-1$ gives $$\sum_{a=0}^{n-1}\bigl(2a + 2(a+1)\bigr) = 2n^2$$ real solutions.

The smallest $n$ with $2n^2 \ge 2012$ is $n = 32,$ since $2 \cdot 31^2 = 1922$ and $2 \cdot 32^2 = 2048.$

Thus, the correct answer is C.