The problem is to maximize the number of positive elements in the vector $Ax$ subject to $x\ge0$, $||x||=1$. The caveat is that $(Ax)_i$ need not be non-negative, i.e., we do not have $Ax\ge 0$. Can this be written as an LP?
Here's what I've got. Clearly, $||x|| = 1$ can be dropped and $x$ can be normalised later. Suppose we "split" $x=y+z$ and write down the following problem:
Maximize the number of positive elements in vector $A(y+z)$
subject to: $Ay\ge0, y+z\ge0$
So now we have a non-negativity constraint on $Ay\ge0$ and we wish to maximize the number of strictly positive elements of $Ay$. (!)
"Splitting" further $y=u+v$ and writing down the following LP after the above link:
Subject to $Av\ge0, 0\le Au\le1, u+v+z\ge0$
Note: $Ay=A(u+v)\ge0$ can be dropped in favour of the above constraints.
Does this solution appear correct? The optimal solution satisfies the constraints: if $x$ is optimal, then so is $cx$ (has the same number of positive elements). This can be used to scale $Ax$ until the least positive element is at least 1. Then split $Ax$ into negative parts ($Az$), set $(Au)_i$ equal to 1 if $(Ax)_i>0$, otherwise $0$, and $Av$ is the rest. [Provided of course such $z, u, v$ exist. Is it always the case? If not, is there a workaround?]