If we think of a row as being the indicator vector for a subset of a set $X$ of $6$ elements, the problem can be stated as, given a family $\mathcal{F}$ of at least $2$ subsets of a set $X$ of $6$ elements, closed to intersection, then there exists some element $x\in X$ such that $x$ is missing from at least half the subsets in $\mathcal{F}$. This rewording has the advantage to offer a natural interpretation to the stuff on the result of that operation on rows.

Here's a sketch of a solution, which I wrote on my phone during a field trip in October. I remember commenting that I was melting. Quote: So uh assume for contradiction that each column has >1/2 ones. So on average there are more ones and zeros and for n>3 that means there will either be a row with exactly four ones or a row with exactly five ones. (Case of 111111 is annoying though). If there's a row 111110 then last column has more than half zeros since anything that ends with 1 has an analogous row ending with 0. If there's a row 111100 then let A be the set of those ending with 01 and B with 10. Note that rows ending with 00 are a "subset" of those ending with 11 like before. So if B>A then last column works; otherwise penultimate column works. To elaborate on the parenthetical note in the first paragraph: note that if 111111 is a row, and all other rows have at most three ones, then there are at most $6+3(n-1) = 3n+3$ ones. This is pretty bad, because we need at least $6\left\lfloor \tfrac{n+1}{2} \right\rfloor$ ones, and so we deduce that all other rows have three $1$'s, and furthermore, $n$ must be odd. But then by multiplying two distinct rows with three ones, we get a row with strictly fewer than three ones, and that's a contradiction.