jmc

Instead of adding up the sum and finding the residue, we can find the residue of each number to make computation easier.

Each group of 6 numbers would have the sum of residues $1+2+3+4+5+0\equiv 15\equiv 3(\mathrm{mod}6)$.

There are $\lfloor \frac{100}{6}\rfloor =16$ sets of $6$ numbers. This leaves the numbers $97,98,99,$ and $100$, which have the residues $1,2,3,$ and $4$. Adding together all the residues, we have $3\cdot 16+1+2+3+4\equiv 58\equiv \boxed{4}(\mathrm{mod}6)$.