Singapore Mathematical Olympiad (SMO)

Consider all the 7 triples $(x,y,z)$, where $x,y,z$ are either $0$ or $1$ but not all $0$. For each $i$, at least one of the numbers $a_i,b_i,c_i$ is odd. Thus among the 7 sums $x{a}_i+y{b}_i+z{c}_i$, 3 are even and 4 are odd. Thus there are altogether $4N$ odd sums. Thus there is choice of $(x,y,z)$ for which at least $4N/7$ of the corresponding sums are odd. (You can think of a table where the rows are numbered $1,2,\dots ,N$ and the columns correspond to the 7 choices of the triples $(x,y,z)$. The 7 entries in row $i$ are the 7 sums $x{a}_i+y{b}_i+z{c}_i$. Thus there are 4 odd numbers in each row, making a total of $4N$ odd sums in the table. Since there are 7 columns, one of the columns must contain at least $4N/7$ odd sums.)