Problem 97

Let $x=y=0$, we get: $f(0+0) = f(0) + f(0)$,
∴ $f(0) = 0$;
Let $y=-x$, we get $f(-x) + f(x) = f(0) = 0$, which means $f(-x) = -f(x)$,
∴ $y=f(x)$ is an odd function;
Since when $x > 0$, $f(x) > 0$,
∴ when $x_1 0$, $f(x_2) - f(x_1) = f(x_2) + f(-x_1) = f(x_2 - x_1) > 0$,
∴ $y=f(x)$ is monotonically increasing on $\mathbb{R}$.
∴ The maximum value of $f(x)$ on $[-2012, -100]$ is $f(-100)$.
Since $f(2) = 4$,
∴ $f(-2) = -4$,
∴ $f(-2-2) = f(-2) + f(-2) = 2f(-2) = -8$, i.e., $f(-4) = -8$,
Similarly, we can get $f(-6) = 3f(-2) = -12$
...,
$f(-2n) = nf(-2)$,
∴ $f(-100) = 50f(-2) = -200$.
∴ The maximum value of $f(x)$ on $[-2012, -100]$ is $\boxed{-200}$.
Hence, the answer is $\boxed{-200}$.