Question: Is it possible to have a matrix $\mathbf B \in \mathbb R^{m\times n}$ such that it satisfies:
$$\mathbf B^T\cdot\mathbf B = \mathbf I\_n$$

Where $\mathbf I\_n$ is the $n\times n$ identity matrix.

Or in other words, is it possible to have a rectangular matrix such that it's transpose is it's left inverse?

If so, what kind of matrix is this? How could you go about constructing such a matrix?

Answer: 
Any matrix of orthogonal column vectors has the property you seek. For two column vectors to be orthogonal, their dot-product has to be 0. This is what AdLibitum's answer is saying. Actually, mathematicians use it as a shorthand for "here's a matrix of orthogonal column vectors".

These orthogonal vectors form an orthogonal basis for the range of the matrix. Since you can orthogonalize any linearly independent matrix, matrices like this are very common. They are also very useful in practice: QR factorization, which decomposes a matrix into an orthogonal and a triangular part, is one of the most important algorithms in numerical computing.

Matrices like these come up in real life all the time. For example, if you take the SVD of a matrix to reveal its range and nullspace, then the range basis vectors and nullspace basis vectors can each form matrices of orthogonal columns.
I assume that $B^T$ has more columns than rows, and that $m=2n$. Then let $C\_{n \times n}$ be a rotation-matrix (=orthogonal) and $D\_{n \times n}=I\_{n \times n}$  

Then let $B^T= 0.5 \cdot [C \quad D]$ being a blockmatrix, then $B^T \cdot B = I\_{n \times n}$