I'm having an hard time trying to figuring out a concrete example of an odddimensional closed manifold that do not admit any contact structure. Can someone provide me with some examples?

3$\begingroup$ Not exactly answering the question since I'm not providing examples, but for a manifold $M^{2n+1}$, it turns out that admitting a contact structure is equivalent to admitting a reduction of structure group to $U(n) \times 1$ (such a reduction is called an almost contact structure), as proved by Borman, Eliashberg, and Murphy. The contact structures they produce for a given almost contact class are overtwisted, meaning they contain some model overtwisted chart. It is a more difficult question to ask when a manifold admits a tight (= nonovertwisted) contact structure. $\endgroup$– KSackelMar 14, 2019 at 0:16
3 Answers
According to a well known result of Martinet, every compact orientable $3$dimensional manifold has a contact structure [2], see also [1] for various proofs. On the other hand we have
Theorem. For $n\geq 2$ there is a closed oriented connected manifold of dimension $2n+1$ without a contact structure.
For $n=2$, $SU(3)/SO(3)$ has no contact structure and for $n>2$, $SU(3)/SO(3)\times\mathbb{S}^{2n4}$ has no contact structure, see Proposition 2.4 in [3].
[1] H. Geiges, An introduction to contact topology, Cambridge studies in advanced mathematics 109.
[2] J. Martinet, Formes de contact sur les variétés de dimension 3. Proceedings of Liverpool Singularities Symposium, II (1969/1970), pp. 142–163. Lecture Notes in Math., Vol. 209, Springer, Berlin, 1971.
[3] R. E. Stong, Contact manifolds. J. Differential Geometry 9 (1974), 219–238.
Although every closed oriented 3 manifold admits infinitely many contact structures, there isa n obstruction to the existence of contact structure on odd dim manifolds of dim$\geq5$.
Suppose that $n>0$ and $\xi= ker \alpha$ is a cooriented contact structure on $Y^{2n+1}$. Then the tangent bundle of $Y$ has a splitting $TY= \xi \oplus \mathbb R$. Then $d\alpha$ is symplectic on $\xi$ and thus $\xi$ admits a compatible complex vector bundle structure[such a splitting called almost complex structure]. This is equivalent to a regudction group to $U(n)\times 1 \subset SO(2n+1,\mathbb R)$. Now lets observe the Chern class of $\xi$ viewed as a complex vector bundle over $Y$. Moreover $c_1(\xi)$ reduced to StiefelWhitney $w_2(Y)\in H^2(Y,\mathbb Z_2)$. So we can say that if $Y$ admits an almost complex structure then $w_2$ admits an integral lift which is equivalent of third StiefelWhitney class $W_3(Y)\in H^3(Y,\mathbb Z)$ vanishes. [Since $W_3$ is the image of $w_2$ under Bockstein homomorphism $\beta$ $\to H^2(Y,\mathbb Z)\to H^2(Y,\mathbb Z_2)\to_\beta H^3(Y,\mathbb Z)\to$.] Thus $W_3=0$ is a necessary condition.
Now if $Y= SU(3)/SO(3)$. Then we have a fibration $SO(3)\to SU(3)\to Y$. By using Steenrod squares, $W_3(Y)$ is the generator of $H^3(Y,\mathbb Z)=\mathbb Z_2$. So this manifold doesnot admit almost contact structure.
Theorem[Borman, Eliashberg,Murphy] There exists a contact structure in every homotopy class of almost contact structure.
Any closed contact manifold is the boundary of an almost complex manifold. (This is written down in detail here.)
In particular, any manifold which is not nullcobordant, cannot carry a contact structure. (The Wu manifold $SU(3)/SO(3)$ is the simplest such manifold.)