For Every Integer 'x'

$f(x)=7x^{19}+19x^7+13ax$ , which can be written as:-

$f(x)=7(x^{19}-x)+19(x^7-x)+13ax+26x$

Now, Fermat once said that $x^{n}-x$ is divisible by $n$ where $n$ is a prime (both $7$ and $19$ indeed are), so $7(x^{19}-x)+19(x^7-x)$ is already divisible by 133.

We need to find two smallest $a$ for which $13ax+26x$ is divisible by $133$ for every integer $x$ . Assuming the worst case, where $133$ doesn't divide $13x$ , it must divide $a+2$ . So the 2 smallest values of $a$ for which this is true are $131$ and $264$ , summing up to $\boxed{395}$ .