Let $k$ be a positive integer. Look at the first row of larger squares in the complement of the blue grid you get from colouring squares corresponding to multiples of $k$ blue. First, we are going to prove that the numbers within the $nth$ larger square from the left add to
$$(2n-1)T_{k-1}^2,$$ where $T_{k-1}$ is the $(k-1)st$ triangular number.
Let's call the square in question $S.$ First note that the numbers within the top row of $S$ are
$$1+k(n-1), 2+k(n-1), 3+k(n-1), ... , \left((k-1)+k(n-1)\right).$$
Adding those up we get the sum
$$1+k(n-1)+2+k(n-1)+3+k(n-1)+...+(k-1)+k(n-1)\right),$$
which we can rearrange to give
\begin{equation}\left[1+2+...+(k-1)\right] + \left[(k-1)k(n-1)\right].\end{equation}
As we mentioned in the main article, there's a formula for the sum of the first $k-1$ integers:
\begin{equation}1+2+...+(k-1) = \frac{(k-1)k}{2}. \end{equation} Substituting this into the left part of expression (1) gives
\begin{equation}\frac{(k-1)k}{2} + (k-1)k(n-1) = \frac{(k-1)k+2(k-1)(n-1)k}{2}=\frac{(k-1)k\left(1+2(n-1)\right)}{2}=\frac{(k-1)k}{2}\left(2n-1).\end{equation}
Now the $(k-1)st$ triangular number is equal to the sum of the first $k-1$ integers, so it's also equal to the right hand side of equation (2). This means that expression (3) is equal to $$(2n-1)T_{k-1}.$$
Now notice that the numbers in the second row of $S$ are 2 times the numbers in the first row, so their sum is equal to
$$2(2n-1)T_{k-1}.$$
The numbers in the second row of $S$ are 3 times the numbers in the first row, so their sum is equal to
$$3(2n-1)T_{k-1}.$$
We can continue in this vein until we come to the sum of the last row of $S$, which is equal to
$$(k-1)(2n-1)T_{k-1}.$$
Adding up the sums of all the rows in $S$ therefore gives
\begin{equation}(2n-1)T_{k-1}+2(2n-1)T_{k-1}+ ... +(k-1)(2n-1)T_{k-1}.\end{equation} Factoring out the $(2n-1)T_{k-1}$ transforms this sum to
\begin{equation}(2n-1)T_{k-1}\left(1+2+3+...+(k-1)\right).\end{equation}
Now the second bracket here is equal to $T_{k-1}$, the $(k-1)st$ triangular number. Therefore our sum (5) is equal to
$$(2n-1)T_{k-1}^2,$$ as we claimed.