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.
Best Answer
The inequality is true.
Lemma Let $\lambda_1 \leq \ldots \leq \lambda_N$ be an increasing sequence of real numbers. Consider the set $\mathcal{A}_k := \{ (x_1,\ldots,x_N) \in [0,1]^N, \sum x_i = k \leq N\}$ of $N$-tuples inside the unit $N$ box, with total sum given by some real number $k$. Then $$ \inf_{x\in \mathcal{A}_k} \sum_{i = 1}^N \lambda_i x_i = \sum_{i = i}^{\lfloor k \rfloor} \lambda_i + (k-\lfloor k\rfloor) \lambda_{\lceil k\rceil} $$
In other words: the lemma says that if you have $N$ different materials costing $\lambda_1\ldots \lambda_N$ per kilogram. You need to buy $k$ kilograms, but you are not allow to buy more than 1 kilogram of each type, then the cheapest way you can satisfy the requirement is to first buy 1 kilo of the cheapest stuff, then another kilo of the second cheapest, and so on until you used up all the money. The proof of the Lemma I'll leave as an easy exercise.
Now we can apply Lubos' observation. Namely that if $X_1, X_2, \ldots X_M$ are $M$ orthonormal vectors in $\mathbb{R}^N$ (with $M < N$), then $$ P = X_1X_1^T + X_2X_2^T + \cdots X_MX_M^T $$ is a projection operator onto some $M$ dimensional subspace. In particular, as a projection, it cannot increase in length! This implies that for each unit vector $v$ $$ v^TPv \leq 1$$ and in particular if we take $v$ to be the standard unit vector $e_i$, this says that the coordinate values $$ \sum_{j = 1}^M (e_i^T X_j)^2 \leq 1$$
On the other hand, we also have the fact that for each fixed vector $X_j$, the normality condition $$ \sum_{i = 1}^N (e_i^T X_j)^2 = 1 $$
So in particular if we define the $N$-tuple $$ Y_i = \sum_{j=1}^M (e_i^TX_j)^2 $$ we have that $\sum_{i=1}^N Y_i = M$, and each individual component $Y_i\leq 1$. So in particular $Y_i \in \mathcal{A}_M$.
Your inequality then follows from the above observation combined with the Lemma.
Note that this also shows why in the case of $N = M$ the inequality must be saturated. The above process converts the problem to the case of the Lemma where you have $N$ types of material, you need to buy $N$ kilograms with no more than 1 kilo of each. So there's only one way of filling the purchase order!