(Z_n, +, other multiplication) can be a field? Is it possible for every $N=p^n, n\in\mathbb{N}^*, p$ prime, to define an operation $\star:\mathbb{Z}_N\times\mathbb{Z}_N\to\mathbb{Z}_N$ such that the structure:
$$(\mathbb{Z}_N, +, \star)$$ will be a field, where the adition is the usual one in the group $\mathbb{Z}_N$?
 A: edit: Taking into account the comments on this answer, I have to add one point: I am assuming in my answer that $1$ is the multiplicative neutral element for the multiplication $\star$. That doesn't need to be true in general. Given the multiplication $\star$ with neutral element $a$, we first show that $1$ is a unit.
For that, use that 
$$a = a\star a = (\sum_{i=1}^a 1)^2 = \sum_{i=1}^{a^2} 1^2 = 1 \star \sum_{i=1}^{a^2} 1.$$
Therefore, let $1^{-1}$ be the inverse of $1$ with respect to $\star$. Then we can define a new multiplication $\star'$ by
$$x \star' y := x \star y \star 1^{-1}.$$
This will give us a multiplication with $1$ as neutral element and by mapping $1 \mapsto a$, both rings are isomorphic (or you get a contradiction with the axioms, e.g. in the case $gcd(a,N) > 1$).
Therefore, we can w.l.o.g. assume that the neutral element is indeed $1$.
Sorry that I forgot about that before, it is not that trivial to get that we can assume $1$ as neutral element, but it is very easy to just assume it without thinking enough about it.

The comments are most likely due to the fact that all finite fields are known and classified, and from that we can easily see that $\mathbb{Z}_N$ with its natural addition will never be one of them, unless $N = p$ is a prime of course (but there we already know which multiplication to take).
As I think that you didn't study this classification yet, let's look for a more elemental proof that this does not work:
Assume that $N$ is not a prime. Then there are numbers $1 < a,b < N$ such that $N = ab$. When we multiply these resp. their residues in $\mathbb{Z}_N$, we can use the distributive law to look at
$$a \star b = (\sum_{i=1}^a 1)\star (\sum_{i=1}^b 1) = \sum_{i=1}^{ab} 1\star 1.$$
Of course you might want to add further steps here, but be sure to only use the axioms for a multiplication, e.g. distributive law, associative law, etc.
Now in the last sum, we sum $N = ab$ ones, so this product needs to be equal to $N$ - no matter how exactly $\star$ was defined. But that is a problem (why?).
Note that in this way, you can even show that the classical multiplication is the only one that makes $\mathbb{Z}_N$ with its normal addition into a ring.
A: As stated in the comments, this is not possible. But you can show even more:

Let $n \in \mathbb N$. Up to isomorphism, there is a unique commutative
  ring with identity with underlying cyclic abelian group $\mathbb Z_n := \mathbb Z/n\mathbb Z$.

Proof.
Since the underlying abelian group has elements of order $n$, the characteristic of the ring is at least $n$. In particular the multiplicative identity $e$ has additive order at least $n$ and subsequently it is equal to $n$. Thus $e$ is co-prime to $n$ and there is an automorphism of the abelian group $\mathbb Z_n$ which sends $e$ to $1$, hence our ring will be isomorphic to the usual ring structure on $\mathbb Z_n$.
A: As several people have said, the answer is no. But there's an interesting reason why. Here, I'll let $\mathbb{F}_p$ denote the "standard" field $\mathbb{Z} / p \mathbb{Z}$ for prime $p$. Let $K$ be a finite field.
Firstly, every field has what's called a characteristic, where $\operatorname{char}(K)$ is the least $n \geq 1$ such that $\sum_{j = 1}^{n} 1_K = 0_K$, where we say that $K$ has characteristic $0$ if no such $n$ exists. But $K$ is finite, so by a counting argument it must have positive characteristic. Moreover, it can be shown that every field of positive characteristic in fact has prime characteristic. So let $p = \operatorname{char} (K)$.
Now I claim that $\mathbb{F}_p$ is a subfield of $K$. Let $\phi : \mathbb{Z} \to K$ be the ring homomorphism $\phi(1_\mathbb{Z}) = 1_K$, which has kernel $p \mathbb{Z}$. Thus by the universal property of quotient morphisms, there exists a unique $\Phi : \mathbb{F}_p \to K$ such that $\phi = \Phi \circ \pi$, where $\pi : \mathbb{Z} \to \mathbb{F}_p$ is the familiar projection. Moreover, we can observe that $\Phi$ is an injection, so we have our embedding of $\mathbb{F}_p$ as a subgroup of $K$. We call $\mathbb{F}_p = \mathbb{F}_{ \operatorname{char} (K) }$ the characteristic subgroup of $K$.
Now, our insightful step is to say that we can treat $K$ as a vector space over $\mathbb{F}_p$, where "scalar multiplication" is just multiplication by $\mathbb{F}_p$. We can construct a basis for the vector space $K$, call it $\{ \theta_1 , \ldots , \theta_n \}$ for some $n$. But this means that $K$ has $p^n$ elements.
Some things that come out from this argument:


*

*It's actually impossible to give a field structure to any set of $N$ elements if $N$ isn't an integer power of a prime.

*If $K$ is a field of order $p$ for a prime $p$, then $K$ is isomorphic to $\mathbb{F}_p$.

*Since every finite integral domain is a field, we can replace every instance of the word "field" in the last two with the word "integral domain".
In fact, a strengthening of 2 is possible, but not shown here: For every positive prime $p$, positive integer $n$, there exists a finite field $\mathbb{F}_{p^n}$ of order $p^n$, and if $K$ is another field of order $p^n$, then $K$ is isomorphic to $\mathbb{F}_{p^n}$. In other words, there exists (up to isomorphism) exactly one field of order $p^n$.
