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. $\endgroup$ – KSackel Mar 14 at 0:16
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.