A grid problem

We will use Vieta's formula for this solution.

Clearly, the numbers $\large b_1, b_2, \dots, b_{2018}$ are the distinct roots to the equation $\large (a_1+x)(a_2+x)\dots (a_{2018}+x)=1$ . By Vieta's formula, we now have $\large a_1+a_2+\dots+a_{2018}= -(b_1+b_2+\dots+b_{2018}), \dots, a_1 a_2\dots a_{2018}-1=b_1 b_2\dots b_{2018}$ .

Now, we will try to find the products of the numbers in each column.

Let us take the first column. The product of the numbers here is $\large (a_1+b_1)(a_1+b_2)\dots(a_1+b_{2018})$ .

From the expressions that we established earlier, we have $\large (a_1-a_1)(a_1-a_2)\dots(a_1-a_{2018})={a_1}^{2018}-{a_1}^{2017}(a_1+a_2+\dots+a_{2018})+\dots+a_1a_2\dots a_{2018}={a_1}^{2018}+{a_1}^{2017}(b_1+b_2+\dots+b_{2018})+\dots+b_1b_2\dots b_{2018}+1=(a_1+b_1)(a_1+b_2)\dots(a_1+b_{2018})+1=0$ (because of $\large a_1-a_1$ ).

Therefore, $\large (a_1+b_1)(a_1+b_2)\dots(a_1+b_{2018})=-1$ .

This is the same for all columns because one of the terms $\large a_k-a_1, a_k-a_2, \dots ,a_k-a_{2018}$ will always be $\large 0$ .

Therefore, the sum of the products of the numbers written in each column is $\large 2018\times -1=-2018$ .