Tuples with all coordinates summing to $3$ 
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 $$\sum_{i=1}^na_i=\sum_{i=1}^nb_i=\sum_{i=1}^nc_i=\sum_{i=1}^nd_i=3.$$ It is known that there exists a partition of $N=\{1,\dots,n\}$ into three sets $N_1,N_2,N_3$ such that $$\sum_{N_1}a_i=\sum_{N_2}a_i=\sum_{N_3}a_i=1.$$ Analogous statements hold for $b,c,d$. Is it always possible to partition $N$ into two sets $X,Y$ so that
  $$\sum_X a_i,\sum_Y b_i,\sum_Y c_i,\sum_Y d_i\geq 1?$$

Without the given condition, this is certainly not true, e.g. $n=1$ and the only tuple is $(3,3,3,3)$. I've thought about putting elements from $N$ into $X$ and $Y$ one-by-one starting with one with the highest $a_i$, then highest $b_i$, highest $c_i$, and so on, but that cannot give more than $3/4$ in the inequality.
 A: I'm gonna use a slightly different notation to take into consideration Ross Millikan's remark.
There are 4 partitions of $N=\{1\dots n\}$ into three sets $\{A_1,A_2,A_3\}$, $\{B_1,B_2,B_3\}$, $\{C_1,C_2,C_3\}$, $\{D_1,D_2,D_3\}$ such that 
\begin{align}
\sum_{A_1} a_i =\sum_{A_2} a_i =\sum_{A_3} a_i =1\\
\sum_{B_1} b_i =\sum_{B_2} b_i =\sum_{B_3} b_i =1\\
\sum_{C_1} c_i =\sum_{C_2} c_i =\sum_{C_3} c_i =1\\
\sum_{D_1} d_i =\sum_{D_2} d_i =\sum_{D_3} d_i =1
\end{align}
If there's a set $A_k$ such that
$$
\sum_{A_k} b_i \le 2\qquad\sum_{A_k} c_i\le 2\qquad\sum_{A_k} d_i \le 2
$$
then $X=A_k$ and $Y={A_k}^C$, since $\sum_{A_k}a_i=1$ and
$$
\sum_{{A_k}^C} b_i \ge 1\qquad\sum_{{A_k}^C} c_i\ge 1\qquad\sum_{{A_k}^C} d_i \ge 1
$$
Otherwise, if no such $A_k$ exists, we can assume without loss of generality
\begin{align}
\sum_{A_1} b_i &> 2 &\sum_{A_2} c_i &> 2 &\sum_{A_3} d_i &> 2\\
\sum_{A_2\cup A_3} b_i &< 1 &\sum_{A_1\cup A_3} c_i &< 1 &\sum_{A_1\cup A_2} d_i &< 1
\end{align}
Now consider $B_1\cap A_1$, $B_2\cap A_1$ and $B_3\cap A_1$. There's at least one set $B_j$ such that
$$
\sum_{B_j\cap A_1} a_i \ge\frac{1}{3}\qquad \sum_{B_j\cap A_1} b_i \le 1
$$
the latter being true for every $B_j$ since $\sum_{B_j\cap A_1} b_i \le \sum_{B_j} b_i$.
Similarly, there exist $C_k$ and $D_l$ such that
\begin{align}
\sum_{C_k\cap A_2} a_i &\ge\frac{1}{3} &\sum_{C_k\cap A_2} c_i &\le 1\\
\sum_{D_l\cap A_3} a_i &\ge\frac{1}{3} &\sum_{D_l\cap A_3} d_i &\le 1
\end{align}
Now take $X=(B_j\cap A_1)\cup(C_k\cap A_2)\cup(D_l\cap A_3)$, notice that it's the union of 3 disjoint sets and
\begin{align}
\sum_X a_i &= \sum_{B_j\cap A_1} a_i + \sum_{C_k\cap A_2} a_i + \sum_{D_l\cap A_3} a_i \ge \frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\\
\sum_X b_i &= \sum_{B_j\cap A_1} b_i + \sum_{C_k\cap A_2} b_i + \sum_{D_l\cap A_3} b_i \le 1 + \sum_{A_2\cup A_3} b_i < 2\\
\sum_X c_i &= \sum_{B_j\cap A_1} c_i + \sum_{C_k\cap A_2} c_i + \sum_{D_l\cap A_3} c_i \le 1 + \sum_{A_1\cup A_3} c_i < 2\\
\sum_X d_i &= \sum_{B_j\cap A_1} d_i + \sum_{C_k\cap A_2} d_i + \sum_{D_l\cap A_3} d_i \le 1 + \sum_{A_1\cup A_2} d_i < 2\\
\end{align}
So, yes, it's always possible to partition $N$ into $X$ and $Y$ as you asked.
