Given that $f: \mathbb{N} \to \mathbb{N}$ is a bijective map and $\sum_{n=1}^{\infty} \frac{f(n)}{n^2} < +\infty$.
Note that, since $f$ is a injective function, $f(n)\geq n$. It follows that
$$\frac{f(n)}{n^2}> \frac{n}{n^2}=\frac{1}{n}$$
By comparison test, $\sum_{n=1}^{\infty} \frac{f(n)}{n^2}$ is always divergent.
Hence, the correct answer is:
(B) zero.