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 $1+3+5+0+2+4+6 \equiv 21 \equiv 0 \pmod7$. 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 $\left\lfloor \frac{199}{14}\right\rfloor=14$ sets of $14$ integers, which is equivalent to $7$ odd numbers in our sum. This leaves $197$ and $199$, which have residues $1+3 \equiv \boxed{4} \pmod7$.
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Find the modulo $7$ remainder of the sum $1+3+5+7+9+\dots+195+197+199.$
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
4
Find the modulo $7$ remainder of the sum $1+3+5+7+9+\dots+195+197+199.$
test