Solution verification: sum at least $4N/7$ times odd 
Problem: Let $a_{j}$,$b_{j}$,$c_{j}$ be whole numbers for $ 1 \leq j \leq N$. Suppose that for each $j$ at least one of $a_{j}$,$b_{j}$,$c_{j}$ is odd. Show that there are whole numbers $r$,$s$ and $t$ such that the sum $$r\cdot a_{j} + s\cdot b_{j} + t\cdot c_{j}$$ (for $ 1 \leq j \leq N$) is in at least $4N/7$ cases odd.

My attempt: Choose $r=s=t=1$. To get an odd sum with 3 numbers for which one is odd for sure you must have that: 2 are even and the other odd or all three are odd. The total amount of possible even-odd combinations for 3 numbers is equal to ${3 \choose 1} +{3 \choose 2} +{3 \choose 3}=7$ (1 odd v 2 odd v 3 odd), so with $r=s=t=1$ we see that in $4/7$ cases, the sum is odd for a fixed $j$. In total we have thus that in $4N/7$ circumstances the sum is odd.
My doubts: So, while reading the question I thought this would seem like a classic pigeonhole principle question and at first I had the intention to go that route, but when I was trying some random cases I came across this with $r=s=t=1$ and with this all the restrictions hold. What do you guys think about my solution?

Let $$u_1 = (0,0,1)$$ $$u_2 = (0,1,0)$$ $$u_3 = (1,0,0)$$ $$u_4 = (0,1,1)$$ $$u_5 = (1,0,1)$$ $$u_6 = (1,1,0)$$ $$u_7 = (1,1,1)$$
and let for each $i\in \{1,2,...,n\}$ define $v_i = (a_i,b_i,c_i)$. ''Connect'' $u_j$ with $v_i$ iff $u_j\cdot v_i\equiv _2 1$ and we count the number of all connections.
It is easy to see that each $v_i$ has degree exactly $4$, so the number of all connections is $4n$. This means that one $u_j$ from the set $\{u_1,u_2,...u_7\}$ must be connected with at least $4n/7$ vectors from $\{v_1,...v_n\}$. So if take $r,s,t$ to be the coordinates of $u_j$ we are done.
