For each Riemannian manifold one can construct the LeviCivita connection. While this connection is unique, we can call a (Riemannian) manifold flat if the LeviCivita connection is flat. However when one tries to think about flatness in terms of pure manifolds (not manifolds plus metric tensor/connection) the natural questions arises:
Question 1 Are there manifolds $M$ with the property that each connection on $M$ is never flat?
Question 2 Is it possible to find a manifold $M$ with the following property: for each Riemannian metric tensor $g$ the corresponding Levi Civita connection is not flat but still there are flat torsionfree connections on $M$?
Question 3 Is it possible to find a manifold $M$ with the following property: each torsion free connection cannot be flat but still there are some flat connections on $M$?

3$\begingroup$ Question 3 is addressed in the first answer here: mathoverflow.net/questions/149492/… $\endgroup$ – Julian Rosen Oct 11 '14 at 13:59

2$\begingroup$ Related to Q1: mathoverflow.net/questions/91852/… $\endgroup$ – Paul Reynolds Oct 13 '14 at 6:24
For question 1 : if a vector bundle admits a flat connection, it comes from a representation of the fundamental group. Thus if $TM$ admits a flat connection and $M$ is simplyconnected, $TM$ is trivial. Hence any simplyconnected $M$ with $TM$ nontrivial gives an example, for instance $M=\mathbb{S}^2$.
I did not understand the first question
Question 1 Are there manifolds with the property that each connection on is never flat?
Because one of course can construct, on any manifold, a connection which is somewhere flat and somewhere not flat.
But of course there are manifolds that do not admit flat affine connections, the simples example is the sphere of any dimension $n>1$. Actually, in dimension 2 the answer (obtained in 1950th , I believe by Kuiper) is as follows: any nonclosed surface admits a flat torsionfree affine connection, and closed surfaces of zero euler characteristic and only them, admit flat torsionfree connection.
As mentioned by Julian Rosen, the answer on the 3rd question is positive and is in the mathoverflow discussion he mentioned
Answer to Question 2 is positive, a simplest example is $S^2 \times S^1$ with the following affine connection: take $R^3\setminus \{0\}$ and quotient it with respect to the group $Z$ acting by $x\mapsto 2 x$. Since the standard flat connection is invariant with respect to the action, the quotient which is $S^2 \times S^1$ carries a flat connection. It has no flat metric of course.
P.S. You question was studied, and the answers I was given were obtained in the framework of the socalled affine structures. Affine structure on a manifold is an atlas such that the transition maps are affine transformations, i.e., are given by the formula $x\mapsto Ax+ B$ for a nondegenerate matrix $A$ and a vector $b$. The existence of an affine structure is equivalent to the existence of a flat torsionfree affine connection.
I will add some detail to the previous answers by abx, Robert Bryant (in the linked question), and V. S. Matveev.
Regarging question 1: A simply connected manifold $M$, admits a flat connection if an only if it is parallelizable.
Indeed, you can paralleltransport a frame (a basis of the tangent space) from some initial point $p$ to every point $q$ along any curve $\gamma$. The fact that the connection is flat implies that the frame that you obtain at the endpoint $q$ won't vary if you perform a continuous variation (homotopy) of the curve $\gamma$; it only depends on the homotopy class. But there's only one homotopy class, so the frame at $q$ is independent of the choice of $\gamma$, and you get a basis of vector fields $X_i$ that are parallel with respect to the connection: $\nabla_YX_i=0$ for all $X_i$ and $Y$. For examples, observe that all the spheres $S^n$ with $n\geq 2$ are simply connected, and the only parallelizable spheres are $S^1$, $S^3$ and $S^7$. If $n$ is even, you can quickly see that there is not even one nonvanishing vector field, by the PoincaréHopf theorem, because the Euler characteristic is 2.
Regarding question 3: You can furthermore prove that spheres of dimenison $n\geq 2$ don't admite symmetric flat connections.
Indeed, if the connection was symmetric (zero torsion), then the vector fields that you obtain in last construction commute: $[X_i,X_j]=0$. Now, since each $X_i$ gives rise to an action of $\mathbb R$ on the manifold $M$, and the actions commute, you get an action of $\mathbb R^n$ by "translations" in the directions of the vector fields $X_i$, the translation by a vector $w\in\mathbb R^n$ being the result of travelling along $M$ in the direction of the vector field $\sum_i w_iX_i$ during one unit of time. The orbits of this action are open sets because the action is locally free (you can act on a point to "wiggle" it in every direction because the vector fields form a basis of the tangent space), and there is only one such orbit because $M$ has one connected component (so the action is transitive). The only compact $n$manifold that admits a locally free, transitive Lie group action of $\mathbb R^n$ is the $n$torus, which is not simply connected. So a compact, simply connected manifold can't have a symmetric flat connection.
Regarding question 2: To see that $S^2\times S^1$ does not admit a flat metric, observe that if it did, then by the CartanHadamard theorem, its universal cover would be diffeomorphic to $\mathbb R^3$. But it's universal cover is $S^2\times\mathbb R$.
As V.S. Matveev observes in his comment below, the CartanHadamard argument can also be used to prove that spheres don't admit a symmetric flat connection.

$\begingroup$ What do you mean by symmetric connection in your answer concernings Question 3? Is it torsionfree? If yes, may be you should slightly correct your answer by saying symmetric flat instead of symmetric? $\endgroup$ – Vladimir S Matveev Oct 12 '14 at 7:46

$\begingroup$ P.S. I have edited my answer a bit without reading yours first, sorry for it, and now it may be an overlap: in the current version of my answer I also claim that any sphere does not admit a flat symmetric connection. Actually, your explanation at the beginning of your answer is (almost) an explanation why any sphere does not admits a flat torsionfree connection, I will explain it in the next comment $\endgroup$ – Vladimir S Matveev Oct 12 '14 at 7:47

1$\begingroup$ Indeed, you explained that for flat torsionsfree connection on a simply connected nmanifold there exist n linearly independent parallel vector fields. Take a Riemannian metric such that at every tangent space in the basis given by these parallel vector fields it is standard euklidean. Since it is constructed by parallel objects, it is also parallel which implies that our (flat torsionfree) connection is its LeviCivita connection. Then, the metric is flat and the argument you used at the end of your answer (with CartanHadamard) can be used also here to show that it is impossible $\endgroup$ – Vladimir S Matveev Oct 12 '14 at 7:49

1$\begingroup$ I corrected the "symmetric flat" thing. I like the idea of using CartanHadamard to prove that spheres don't admit torsionfree flat connections. This saves the task of proving that only toruses admit free $\mathbb R^n$ actions, which is something I didn't do in my answer. $\endgroup$ – Marcos Cossarini Oct 13 '14 at 11:37

2$\begingroup$ It is almost an offtopic but the statement that the ntori are the only closed nmanifolds that admit a locally free $R^n$ action is standard in the theory of integrable systems and a good proof can be found in the Arnold's ``Mathematical methods of classical mechanics''. There are of course proofs in other sources, say in Berger's ''Geometry''. Though the statement is intuitively expected, the proof is still slightly tricky and may require some time. $\endgroup$ – Vladimir S Matveev Oct 13 '14 at 17:51