Some $4$-tuples of positive real numbers $(a_1,b_1,c_1,d_1),\dots,(a_n,b_n,c_n,d_n)$ are given, with all $a_i,b_i,c_i,d_i\leq 1$. Can we always partition $\{1,2,\dots,n\}$ into two subsets $X,Y$ so that $$1+\sum_Xa_i\geq \sum_Ya_i\text{ and } 1+\sum_Xb_i\geq \sum_Yb_i$$
and $$\sum_Xc_i\leq 1+\sum_Yc_i\text{ and } \sum_Xd_i\leq 1+\sum_Yd_i?$$
It could be that some inequalities need to be exactly tight, for example if $n=1$ and the only tuple is $(1,1,1,1)$. This also means that the number $1$ in the inequalities cannot be replaced by a smaller number.
Mike Earnest shows below that the partition is possible if we replace the $1$'s by $3$'s. What is the best possible value between $1$ and $3$?
Best Answer
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 $x_i=0$.
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.