1's

Note that the first sum is a string of $1007!+2$ 1's, and the second is a string of $1009$ 1's.

We can group the 1's in the first sum together into strings of $1009$ , in however fashion we like. Thus, we start from the left, taking off $1009$ 1's at a time. This is equivalent to taking $1007!+2 \pmod{1009}$ , trying to find how many digits are left at the end.

Because

$1008! = (1009-1)! \equiv -1 \pmod{1009}$

due to Wilson's theorem, and

$1008! = (1009-1) \times 1007! \equiv -1007! \pmod{1009}$

it is then the case that

$1007!+2 \equiv 1+2 \equiv 3 \pmod{1009}$

Since there are $3$ 1's left over, our remainder is $\boxed{111}$ .

I use the fact that if $n=dq+r$ for $0\leq r <d$ , then $\frac{x^n-1}{x^d-1}=(x^d-1)f(x)+\frac{x^r-1}{x^d-1}$ , for some polynomial $f(x)$ .

I also use $\displaystyle \sum_{k=0}^{n-1} x^k =\frac{x^n-1}{x-1}$ .

Thus, $\frac{\displaystyle \sum_{k=0}^{1007!+1} 10^k}{\displaystyle \sum_{k=0}^{1008} 10^k}= \frac {10^{1007!+2}-1}{10^{1009}-1}=(10^{1009}-1) \cdot f(10) +\frac{10^r-1}{10^{1009}-1}$ .

It turns out that $1009$ is prime, so by Wilson's theorem, $1008!=-1$ mod $1009$ , so $1007!+2=3$ mod $1009$ . Therefore, $r=3$ , so $\frac{10^r-1}{10^{1009}-1}=\frac{10^3-1}{10^{1009}-1}=\frac{10^2+10+1}{\displaystyle \sum_{k=0}^{1008} 10^k}$ ,

So $\displaystyle \sum_{k=0}^{1007!+1} 10^k=10^2+10+1=111$ mod $\displaystyle \sum_{k=0}^{1008} 10^k$