In order to show something, I would like to have this strange side result, which seems obvious yet surprisingly I cannot find a way to show it "rigoriously".
Suppose we have a sequence $(u_n) \subset H^1_0(\Omega)$, such that $$ n \int u_n^-v \to C(v) $$ as $n \to \infty$. Here $(\cdot)^-$ denotes the non-positive part of the original function, $v \in H^1_0(\Omega)$ is a test function and $C(v)$ is just something depending on $v$ (not relevant here, the point is that this is bounded, which means that the integral with the sequence is of order $O(n^{-1}))$. From that we can infer that in the limit $u$ (and it was shown we can pass to the limit etc.) will be non-negative, as this is the only way to avoid the blow-up. Can we thus conclude that $$ ||n\,u_n^-||_{L^2} \leq C_2 ? $$
It seems perfectly obvious, yet any standard inequality does not allow me to conclude it. In particular, playing with $$ \int n^2\, u_n^- \,u_n^- $$ does not seem to yield anything fruitful. I suppose the difficulty lies in the fact that this order of $O(n^{-1})$ is 'hidden' in the integral and it cannot be extracted that easily. Or, and this far more likely, I am be missing something obvious. Any help will be greatly appreciated!
No, you don't have $nu_n^-$ bounded in $L^2$. You would have this if you knew that the sequence $nu_n^-$ is weakly convergent in $L^2$. But the latter requires testing against arbitrary $v\in L^2$. Your test functions are in $H_0^1$, which (by Sobolev embedding) are in some $L^q$ with $q>2$.
A counterexample can be obtained as follows. Let $q$ be as above and choose $p$ so that $1/p+1/q=1$. Pick a nonnegative compactly supported function $f\in L^p\setminus L^2$; this is possible because $p<2$. Specifically, it should smooth except for a singularity like $|x-x_0|^{-\alpha}$.
Let $u_n^- = \min(f,n)/n$; this function is in $H^1$. Since $nu_n^- \to f$ in $L^p$, we have $\int nu_n^-v$ converging for every $v\in L^q$. On the other hand, $\|nu_n^-\|_{L^2}\to \|f\|_{L^2}= \infty$.