I have to prove the negation of this statement: $$\forall a,b,c\in\mathbb{Z}{\;if\;a\;|\;b\} $$
But the fact that there is a "and" is very disturbing. I think that I am missing something because my "version" of the contraposive is obviously wrong. (take a = 3 , b = 7, c = 4)
Your statement is equivalent to $$\forall a,b,c\in\mathbb Z\left( \text{ if } a\mid b+c, \text{ then either } a\mid b\text{ and } a\mid c \text{ or } a\not\mid b \text{ and } a\not\mid c\right)$$
Assume $a\mid b+c$. Then $b+c=ak, k\in\mathbb Z$. Now check two cases, which are exhaustive:
$1)$ $a\mid b$. Then $b=am, m\in\mathbb Z$ and so $b+c=am+c=ak\iff c=a(k-m)$.
Thus $\exists n\in\mathbb Z \left(c=an \right), a\neq 0$, which means $a\mid c$ by the exact definition of divisibility.
$2)$ $a\not\mid b$. Then $b=am-r, 0<r<|a|$ and $b+c=am-r+c=ak\iff c=a(k-m)+r$.
Since $0<r<|a|$, we know that $r$ is the remainder of $c$ when $c$ is divided by $a$.
Since the remainder is not zero, we have that $a\not\mid c$.