The sum of the products in the top row would just be $(1+2+3+4+5+6+7+8+9)=45$
Then the next row would be $(2+4+6+8+10+12+14+16+18) = 2\times45 = 90$
So the top two rows sum to $(1+2)\times 45 = 135$
Then it becomes obvious that the full sum of the products is the product of the sums, ie. $45\times45 = 2025$
Assuming you do not need rigorous, mathematical justification, you can simply do it in the following way:
First, find $n = \dfrac{10^9}{7}=142857142$. Then there will be in total of $ T_n = \dfrac{n(n+1)}{2}$ triangular block of consisting of $1's$. As you demonstrated in your code, each of those block has exactly $T_6 = 21$ $1's$, so there are $N = 21T_n$
numbers that are divisible by $7$ in the first $10^9$ rows of the Pascal triangle. Once you have found this, the answer to actual problem will simply be $1+2+...+10^9 - N$, obviously.
Last but not least, notice that how the only rows that has no $1's$ are the rows numbered $6, 13, 20,...$. In fact, this pattern continues and $10^9$ th row will precisely be another zero row because $10^9= -1\mod 7.$ Therefore, you do not need to worry about half triangle appearing at the very bottom. I am certain that the number $10^9$ was chosen for this particular purpose.
One can prove all these assertions mathematically, but you need at least Lucas's theorem, or some equivalent machinery which in general assumes a very decent knowledge of number theory. The author of that article you mentioned did a very poor job of attempting explain the reasoning.
Best Answer
Not extremely enlightening but I ran this python code to find an answer of $436$, (and a max of $947$).