jmc

The numbers that are shaded are the triangular numbers, which are numbers in the form $\frac{(n)(n+1)}{2}$ for positive integers. They can also be generated by starting with $1$, and adding $1,2,3,\mathrm{4...}$ as in the description of the problem. Squares that have the same remainder after being divided by $8$ will be in the same column. Thus, we want to find when the last remainder, from $0$ to $7$, is found. So, instead of adding $1,2,3,4,5,6,7,8,9,\mathrm{10...}$, we can effectively either add $1,2,3,4,5,6,7,0,1,2,\mathrm{3...}$ or subtract $7,6,5,4,3,2,1,0,7,6,\mathrm{5...}$ if we are only concerned about remainders when divided by $8$. We will pick the number that keeps the terms on the list between $1$ and $8$. We get: $1$ $1+2=3$ $3+3=6$ $6-4=2$ $2+5=7$ $7-2=5$ $5-1=4$ $4+0=4$ $4+1=5$ $5+2=7$ $7-5=2$ $2+4=6$ $6-3=3$ $3-2=1$ $1-1=0$ Finally, a term with $0$ is found, and checking, all numbers $1$ through $7$ are also on the right side of the list. This means the last term in our sequence is the first time that column $8$ is shaded. There are $15$ terms in the sequence, leading to an answer of $\frac{15\cdot 16}{2}=120$, which is choice $\boxed{120}$.