First of all, one can shift the set $\mathcal{J}$ so that the condition $x_i \leq 1$ can be replaced by $x_i \geq 0$.
Then one should note that if there is a vector $x^1$ in $\mathcal{J}$ withLet $x^1_i > 0$ and a vector$M_i$ be the maximal absolute value of $x^2$$x_i$ in $\mathcal{J}$ with $x^2_j> 0$, then we also have a vector $x^3\in \mathcal{J}$ with $x^3_i>0$, $x^3_j>0$ given by any convex combination ofcan introduce binary variables $x^1$$v_i$ and write the problem as:
$\max \sum_{i\in S} v_i$
$M(v_i - 1) \leq x_i$ $x^2$$\forall i \in S$.
So we can solve the Problem $P_i$: $\max x_i$, $x \in \mathcal{J}, x_j \geq 0\, \forall j$ for every $i$ and use a convex combinationThe LP-relaxation of solution vectors to solvethis description is probably poor because of the original problemlarge constant $M_i$.