Can the additive and the multiplicative inverses to an element in $Z_p$ be the same? Does it contradict the axioms of a field? I think not.
If not so there need to be $a \in Z_7$ and so $3+a=0$ and $3\cdot a=1$ but I can not find this $a \in Z_7.$
 A: If $a+b=0$, then $b=-a$. If simultaneously $1=ab$, then $1=-a^2$ which is equivalent to $a^2=-1$. This is possible in $\Bbb{Z}_p$, if and only if $p=2$ or $p\equiv1\pmod4$. 
So you're right. This cannot happen in $\Bbb{Z}_7$.
A: Absolutely!  I wrote a Python program a while ago to generate some examples for general quotient rings $\mathbb{Z}_n$.  Here are some:
$2$ and $3$ $\pmod{5}$
$3$ and $7$ $\pmod{10}$
$5$ and $8$ $\pmod{13}$
$4$ and $13$ $\pmod{17}$
Anyone interested can find my Python code here.  In general, let's suppose a pair of such elements exists in the ring $\mathbb{Z}_m$.  They will satisfy the system $x+y \equiv 0 \pmod{m}$ and $xy \equiv 1 \pmod{m}$, and from this we discover that such a pair exists if and only if there exists an $x \in \mathbb{Z}_m$ such that $x^2 \equiv -1 \pmod{m}$.    

To expand on your specific question regarding $\mathbb{Z}_p$ when $p$ is prime, note that if there is an element $x$ such that $x^2 \equiv -1 \pmod{p}$, then in particular $x^4 \equiv 1 \pmod{p}$.  That is, there must exist an element of order $4$ in the multiplicative group $\mathbb{Z}_p^\times$.  Now, it is a theorem that $\mathbb{Z}_p^\times \cong \mathbb{Z}_{p-1}$ when $p$ is prime.  Finally, since $\mathbb{Z}_{p-1}$ is cyclic, it will contain an element of order $4$ $\iff$ $4|(p-1)$.  So this phenomenon occurs in $\mathbb{Z}_p \iff$ $p \equiv 1 \pmod{4}$.
