Set of size 3 is at least the score Given $5$ triples of nonnegative numbers $(a_{11},a_{12},a_{13}),(a_{21},a_{22},a_{23}), \dots,(a_{51},a_{52},a_{53})$. Let $A=\{1,2,3,4,5\}$. For each $1\leq j\leq 3$, define the $j$th score $s_j$ to be 
$$s_j=\max_{B\subseteq A}\left[\sum_{i\in B} a_{ij}\text{ such that }\sum_{i\in B} a_{ij}\leq \sum_{i\in A\backslash B} a_{ij}\right]$$
Can we always pick a set $C\subseteq A$ of size $3$ such that $\sum_{i\in C}a_{ij}\geq s_j$ for all $1\leq j\leq 3$? (It is not hard to show that any set $C$ of size $4$ satisfies this property.)
Example: The triples are $(16,4,9), (2,6,8), (18,13,1), (19, 13, 6), (3,15,1)$. 
To compute the first score, we look at the set $\{16,2,18,19,3\}$ and find that $24=2+3+19$, which is less than $16+18$, is the score.
To compute the second score, we look at the set $\{4,6,13,13,15\}$ and find that $25=4+6+15$, which is less than $13+13$, is the score.
To compute the second score, we look at the set $\{1,1,6,8,9\}$ and find that $11=1+1+9$, which is less than $6+8$, is the score.
To satisfy the condition, we may pick the 1st, 3rd, and 4th triples, yielding scores of $53, 30, 16$, respectively.
 A: The answer is yes.
Fix $j$ for now and suppress it in the notation; so we consider a quintuplet $(a_1,\cdots,a_5)$ and its score $s$. Call a subset $C\subseteq A$ big if $\sum_{i\in C}a_i\geq s$, and let $\sigma=\frac12\sum_{i\in A}a_i$. Note that $s\le\sigma$, so every $C\subseteq A$ that sums to at least $\sigma$ is big. Assume $a_1\le\cdots\le a_5$ for now. 
Either a) $a_1+a_2+a_5\ge\sigma$, and then all $C\subseteq A$ with $5\in C$ are big, or b) $a_3+a_4\gt\sigma$, and then all $C\subseteq A$ with two elements greater than $2$ sum to more than $\sigma$.
In case a), this leaves $4$ subsets of $3$ elements that might not be big, the ones that don't contain $5$.In case b), the score is $a_1+a_2+a_5$, which  leaves only $2$ subsets of $3$ elements that are not big: $\{1,2,3\}$ and $\{1,2,4\}$.
So of the $10$ subsets of $3$ elements, in case a) up to $4$ are not big and in case b) $2$ are not big.
Now consider the three quintuples for $1\le j\le3$. Now we can no longer fix the order, since the three quintuplets might be in different orders and we have to find a single $C\subset A$ for all of them.
Unless at least two of the quintuples are of type a), less than $10$ of the subsets of $3$ elements are excluded, so we can find one.
If exactly two of the quintuples are of type a) and one is of type b), then the one of type b) excludes at most $2$ of the $3$ subsets of $3$ elements that contain the greatest elements of the two quintuples of type a), so there's at least one left that we can pick.
If all three quintuples are of type a), we can pick the subset containing their three greatest elements.
That exhausts all the cases, so we can always find $C\subset A$ as desired.
