Solution:

Consider a graph with vertices the lines of the table and edges that join two vertices if the respective lines are different exactly in one position. Write on the edges the respective elements that are different.

Assume that the given statement is not true. Then the graph has $n$ vertices and $n$ edges. It is not difficult to prove by induction on $n$ that there is a cycle in the graph. We may suppose that this cycle is $A_{1} A_{2} \ldots A_{k}$.

Starting from $A_{1}$, we remove $x_{1}$ to get $A_{2}$, then remove or add $x_{2}$ to get $A_{3}$, etc. When we turn back to $A_{1}$, we obtain a second copy of $A_{1}$ which does not contain $x_{1}$ (any edge contains no $x_{1}$), a contradiction.