Let $S_m=\{-1,1\}^m$ be the hypercube of signs. Define the set of "admissible weights" $W_m$ as the subset of $\{w\in\mathbb{R}_+^m : \|w\|^2=m\}$ with a "support property" of the corresponding threshold function $f_w: S_m\to \mathbb{R}$ given by
$$ f_w(s_1,\ldots,s_m)=s_1 w_1+\cdots s_m w_m\;. $$
For $w\in W_m$ to be admissible, then $f_w(s)\ne 0$ for all $s\in S_m$ and for all $i\in\{1,\ldots,m\}$ there exists some $s^*\in S_m$ such that the sign of $f$ is flipped when only the sign of $s^*_i$ is flipped:
$$ \mathrm{sgn}{f(\ldots,s^*_i,\ldots)} = -\mathrm{sgn}{f(\ldots,-s^*_i,\ldots)}\;.$$
It's a "support property" because clearly only weights with support $m$ are admissible.
Here is the optimization problem:
Characterize the weights $w^*\in W_m$ that maximize the "margin"
$$ \mu_m=\min_{s\in S_m} |s\cdot w^*|\;. $$
Here's what I know so far. We start at $m=3$ since $W_2$ is the empty set.
$m=3: \qquad w^*=(1,1,1)\qquad\mu_3=1$
$m=4: \qquad w^*=\sqrt{4/7}\;(1,1,1,2)\qquad\mu_4=\sqrt{4/7}$
$m=5: \qquad w^*=(1,1,1,1,1)\qquad\mu_5=1$
I suspect this odd-even pattern continues, of unique margin maximizers at odd $m$ and $m$-fold sets at even $m$.
Edit:
Boolean threshold functions (BTFs) are miniature models of classification, where the points of a hypercube are partitioned into two sets by a hyperplane. The standard representation of a BTF is like the activation functions of machine learning:
$$ b_w(s)=\mathrm{sgn}(w\cdot s)\;. $$
There's a continuum of $w$'s that represent the same BTF, so it's natural to ask which is best. I think everyone would agree that you want to maximize the margin, the interval over which the "decision" takes place:
$$ \mu=\min_{s\in S_m} |w\cdot s|\;. $$
Geometrically, this is proportional to the width of the thickest slab that separates the two sets of hypercube points. This shows the thickest slabs that arise for $m=3$ :
The corresponding weights (with the same normalization as above) are:
$$ (\sqrt{3},0,0) $$ $$ (1,1,1) $$
The thickest one, with $\mu=\sqrt{3}$, is not a good choice for a 3-input BTF since it can be implemented as a 1-input BTF (whose only function is to flip or not-flip the sign of the single input (like a NOT gate). My "support property" of admissible weights restricts the BTF-support to be $m$. When changing the sign of one input ($s_i$ above) has no effect on the value of the BTF, then the corresponding weight can just as well be zero (and so $w$ no longer has support $m$).
Here is what happens for $m=2$. Without loss of generality, $w=(w_1,w_2)$ and $w_1>w_2$ (when $w_1=w_2$ there exists an $s$ such that $w\cdot s=0$). But the sign of $w_1 s_1+w_2 s_2$ is now independent of $s_2$ (no different than the 1-input BTF we get by setting $w_2=0$).
BTFs have been around forever, but my conjectured relationship between support and a bound on the margin seems new.
