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 7 numbers would have the sum of residues $1+3+5+0+2+4+6\equiv 21\equiv 0(\mathrm{mod}7)$. Since we only have odd numbers in the sum, every $7$ odd numbers is $14$ integers. Because every group has a residue of $7$, we can ignore them.

There are $\lfloor \frac{199}{14}\rfloor =14$ sets of $14$ integers, which is equivalent to $7$ odd numbers in our sum. This leaves $197$ and $199$, which have residues $1+3\equiv \boxed{4}(\mathrm{mod}7)$.