Easy to use induction to get $f(n) = 100n$ for all $n\in \mathbb{Z}$. Then $$\sum_{k=1}^{10} f(k!) = 100 \cdot \sum_{k=1}^{10} k! = \boxed{403791300}.$$

$\sum_{k=1}^{10} f(k!)=f\Big(\sum_{k=1}^{10} k! \Big)=f(4037913)=4037913.f(1)=403791300$

Since $f(x+y)=f(x)+f(y)$, we have $f(x)=kx$ for all $x\in\mathbb{Z}$ (Cauchy's FE), and since $f(1)=100$, $k=100$. So the function is $f(x)=100x$. Thus, the answer is $(1!+2!+3!+4!+5!+6!+7!+8!+9!+10!)\cdot100=\boxed{403791300}$.