What is an example of a manifold that is not smooth?

In light of OP's response, here is a construction borrowed from my class notes. However, to show that this manifold does not admit any smooth structure will not be discussed in my answer. The interested reader is welcome to check the other class note taken by Prof. Somanth Basu.

Kervaire claimed that there exists a ten dimensional closed topological manifold which does not support any smooth structure $K^{10}$. In terms of embedding, this means that although by a slight modification of above argument, $K^{10}$ can be topologically embedded into a subset of Euclidean space. There does not exist a smooth submanifold, $M^{10}\subset R^{m}$ such that $M^{10}\cong K^{10}$ homeomorphically. In particular we cannot define a tangent plane globally.

Consider a triangle, which is a topological manifold that is homeomorphic to the circle but not differeomorphic as it has corners. And so is a pentagon. For the $K^{10}$ example, if I remove a point it would be smoothable.

Kervaire's example admits the following relation: $$ H_{*}(K^{10})=H_{*}(\mathbb{S}^{5}\times \mathbb{S}^{2}) $$

We try to approach this by considering $\mathbb{S}^{1}\times \mathbb{S}^{1}-\mathbb{D}^{2}$. I claim that this is homotopically equivalent to $$ \mathbb{S}^{1}\wedge \mathbb{S}^{1} $$ And I can prove this by gluing two annulus orthogonally, if I fill in a disk it would become a torus. Similarly we can consider $\mathbb{S}^{5}\times \mathbb{S}^{5}-\mathbb{D}^{10}$. This would be the same as $$ \epsilon^{5}_{\mathbb{S}^{5}}=\mathbb{S}^{5}\times \mathbb{D}^{5}\cup \mathbb{S}^{5}\times \mathbb{D}^{5} $$ such that we map it by $$ \mathbb{D}^{5}\times \mathbb{D}^{5}\rightarrow \mathbb{D}^{5}\times \mathbb{D}^{5}:(x,y)\rightarrow (y,x) $$ We now cinsider a more general case in which one bundle is the disk bundle of dimension $5$ over the sphere. Another candidate we may consider is the tangent bundle. Since $\mathbb{D}^{5}\subset \mathbb{S}^{5}$, we have $$ N^{10}=\mathbb{DTS}^{5}\bigcup \mathbb{DTS}^{5} $$ where we glue the two disk subbundles $$ \mathbb{D}^{5}\times \mathbb{D}^{5}\rightarrow \mathbb{D}^{5}\times \mathbb{D}^{5} $$ We can claim that the boundary of $N^{10}$ is homeomorphic $\mathbb{S}^{9}$. We know that ananlogous to $n=1$ case we have $$ N^{10}\sim \mathbb{S}^{5}\mathbb{S}^{5} $$ Then we have $$ H_{*}(\partial N)=H_{*}(\mathbb{S}^{9}) $$ by long exact sequence of homology.

Then by Whitehead's theorem we can prove that they are homotopically, and by Smale's theorem we can show that $$ \partial N\cong \mathbb{S}^{9} $$ homeomorphically. After filling a disk $\mathbb{D}^{10}$, we call $$ K^{10}=N^{10}\cup \mathbb{D}^{10} $$ It now took some effort to show that this bundle has no smooth structure.

Remark The long exact sequence in homology calculation only works for $n$ is odd dimensional, for $n=8$ this would fail.

Just a little research gave an answer. http://en.wikipedia.org/wiki/Differentiable_manifold#Relationship_with_topological_manifolds.

Also, most people cannot think in dimension $\geq 4$, so that explains why you couldn't think of a non-smoothable manifold.