If a topological space is homeomorphic to a smooth manifold, then will it be a smooth manifold? If I have already known a topological space $N$ is homeomorphic to a smooth manifold $M$ then will it be a smooth manifold? The atlas of $N$ is the preimage of the atlas of $M$ and the coordinate map is the composition of homeomorphism composites the coordinate map?
 A: Let me emphasize something that is perhaps not readily apparent in the other answers. "Smooth" is not something that a topological space is. You can't formally say "this topological space is a smooth manifold." It is not a property of a topological space. It is extra structure that you add on top of the already existing space.
Indeed, you have to choose an atlas such that all the coordinate change maps are smooth. If you choose an atlas at random, chances are, it won't be smooth, even if you know otherwise that the space is homeomorphic to a smooth manifold.
However there is a property here: can the space be attributed a smooth manifold structure? That is a possible property, it's a yes/no question. In this case the answer is yes! As the other answer points out, you can use the homeomorphism in a manner called transport of structure to choose an atlas that you know will be smooth. But there may be others! For example the $7$-sphere $S^7$ has famously several different, non-equivalent smooth structures (the ones other than the standard one are called exotic).
Therefore it's important to distinguish property and structure. When you ask "Is this space a smooth manifold?" you're implicitly saying that "being a smooth manifold" is a property, which it isn't. And in this particular case it may help you clear up the question: when you wonder "can this space be made into a smooth manifold", then you see immediately that you need to add some stuff on top of what you already have, namely, an atlas. You can't just look at the space and ask yourself "is this smooth" in the same way that you look at a car and ask "is this red". And since you now know what you have to do, it's certainly easier to actually do it. (See also this other answer of mine or this nLab article.)
A: Yes, you can transport the structure; see jabo's answer. However, the smooth structure you induce in this way may not be the natural one. I have in mind the example of a square in $\mathbb R^2$. The square is homeomorphic to a circle, which is a smooth manifold. However, the square is not a smooth submanifold of $\mathbb R^2$, because it has corners. 
Therefore, the smooth structure you import from the circle is not the natural one that you expect on subsets of $\mathbb R^2$.

Let me try to illustrate the main point with a toy model from linear algebra. Consider the vector space $(\mathbb R^2, +, \cdot )$. One may ask: does there exist a vector space structure on
$$A=\left\{ (x, y)\in \mathbb R^2\ |\ x=1\right\},$$
which is a subset of $\mathbb R^2$? The answer is affirmative, because the map 
$$
\phi\colon A\to \mathbb R,\qquad \phi(x, y)=y, $$ 
is a bijection and so you may transport the vector space structure of $\mathbb R$ on $A$.$^{[1]}$ 
However, a more natural question would be whether $A$ is a vector subspace of $\mathbb R^2$. And this is not true, as you surely know.
Summarizing, the structure you transported on $A$ is completely unrelated to the structure of the bigger space $\mathbb R^2$, and so it is less likely to be useful. This is a general fact: if you transport structures, you might well end up with something crazy and potentially useless.

[1] This means that if $(x, y), (x',y')\in A$ you can define $$(x,y)+_\phi(x', y')=\phi^{-1}(y+y'),$$
and if $\lambda\in \mathbb R$ you can define
$$\lambda\bullet_\phi(x, y)=\phi^{-1}(\lambda y)$$
and $(A, +_\phi, \bullet_\phi)$ is a vector space.
A: Sure this works and is sometimes called transport of structure.
