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 $xa_i + yb_i + zc_i$, 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 $1, 2, \ldots, N$ and the columns correspond to the 7 choices of the triples $(x, y, z)$. The 7 entries in row $i$ are the 7 sums $xa_i + yb_i + zc_i$. 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.)