Partition of 4-tuples

I can solve a weaker version of the problem, where instead the "1"s in the target inequalities are replaced with "4"s. Perhaps the work here can help someone else solve the problem.

For each $1\le i \le n$, let $\def\v{{\bf v}}\v_i$ be the vector $(a_i,b_i,-c_i,-d_i)$.

The goal is to solve the following inequality in $n$ variables $x_i$, each equal to $\pm1$, $$ \sum_{i=1}^n x_i \v_i \le (4,4,4,4)\,\qquad x_i\in \{-1,+1\}\tag 1 $$ This is an inequality of vectors, which means the inequality must hold for each component. Setting $x_i=+1$ corresponds to putting $i\in Y$, while $x_i=-1$ means putting $i\in X$.

Let us relax the constraint on the $x_i$, allowing them to take any value in the interval $[-1,1]$, but demanding equality: $$ \sum_{i=1}^n x_i\v_i={\bf 0}\qquad -1\le x_i \le 1,\tag2 $$

where ${\bf 0}$ is the zero vector. This has a solution; namely, set each $x_i=0$.

I claim there is a solution where at least $n-4$ of the variables $x_i$ are equal to $\pm 1$.

Proof of claim: Suppose there is a solution where the variables $x_1,x_2,x_3,x_4,x_5$ are all in $(-1,1)$. Since the vectors $v_1,\dots,v_5$ are linearly dependent, there exist coefficients $c_i$ not all zero so $\sum c_i v_i=0$. Let $c_j=0$ for $j>5$. Then, for any $t\ge 0$, it is also true that the values $x_i+tc_i$ solve the equality in $(2)$: $$ \sum_{i=1}^n (x_i+tc_i)\v_i={\bf 0} $$ Now, letting $t$ increase from $0$ towards $\infty$ until the first time when some $x_i+tc_i=\pm1$, we obtain a different solution $x_i+tc_i$ to $(2)$ with one more variable equal to $\pm 1$. $\square$

So far, we have found a solution to $(2)$ where all but four of the variables are equal to $\pm1$. WLOG these four variables are $x_1,x_2,x_3,x_4$. Now, instead set $$ x_1'=1,\quad x_2'=1,\quad x_3'=-1,\quad x_4'=-1 $$ Since $x_1'-x_1\le 2$ and $x_2'-x_2\le 2$, it follows that $$x_1'a_1+x_2'a_2+x_3'a_3+x_4'a_4+\sum_{i=5}^n x_i a_i\le (2+x_1a_1)+(2+x_2a_2)+x_3a_3+x_4a_4+\sum_{i=5}^n x_i a_i=4+0$$ so the equality in $(1)$ is satisfied.