Briefly, have the following problem: \begin{equation} \sum_{i = 0}^n a_i \ (max [ F_i( \bar x ), 0 ] )^2 \rightarrow min, \\\\ s.t.\\\\ A \bar x \leq b \end{equation} where $ F( \bar x ) $ is a linear function, $a_i \gt 0$, $n$ is huge comparing to the size of $x$.
It is possible to write an equal Quadratic Programming problem, such as
$$ \sum_{i=0}^n a_i \ ( G_i )^2 \rightarrow min \\\\ s.t. \\\\ G_i \geq {\bf 0}, \quad i = 0..n \\\\ G_i \geq F_i( \bar x ) \quad i = 0..n \\\\ A \bar x \leq b $$
which can be solved very efficiently with an appropriate numerical method.
Unfortunately in my particular case such conversion doesn't work: it adds a lot of new restrictions, and that appropriate numerical method doesn't converge.
I tried to figure out another equal QPP, which adds fewer new constraints, but nothing came across my mind. Is there another way?
This is not really an answer. I just want to say that your optimization problem can be converted into a linear programming problem:
If the minimum found is $m$ and the minimizer is $x_0$, then the minimum for your original problem is $\max(m,0)^2$ and the minimizer is $x_0$.
Reason: If $F(x_0)=m<0$, then $\max(F(x_0), 0)^2=\max(m, 0)^2=0$, which is the least possible value of $\max(F(\bar{x}), 0)^2$ over the whole space. Hence $x_0$ is a feasible and global minimizer.
If $F(x_0)=m\ge0$, then $F(\bar{x})\ge F(x_0)\ge0$ for every $\bar{x}\in D=\{\bar{x}: A\bar{x}\le b\}$. Hence $\max(F(\bar{x}), 0) = F(\bar{x})\ge0$ for every $\bar{x}\in D$. Therefore $$\min_{\bar{x}\in D} \max(F(\bar{x}), 0)^2 = \min_{\bar{x}\in D} F(\bar{x})^2 = \left(\min_{\bar{x}\in D} F(\bar{x})\right)^2 = m^2 = \max(m,0)^2.$$