Multiplication and the vector space $\mathbb R^n$ Is it possible to define multiplication of two vectors in the vector space $\mathbb R^n$ such that the cancellation laws hold?
Of course, for $n=1,2$ (real numbers and complex numbers) it is possible, but I am asking for the general case.
Related: Multiplication in $\mathbb{R}^n$
 A: The answer will depend on what other properties you wish the vector space to have.
By Frobenius's theorem, there exist exactly $3$ finite-dimensional unital associative division algebras$^\dagger$ over the reals up to isomorphism.  These correspond to the real numbers $\Bbb R$, complex number $\Bbb C$, and the quaternions $\Bbb H$.  The quaternions are not commutative.
By Hurwitz's theorem, if we remove the associative property, there exist such algebras only for $n=1,2,4,8$.  These correspond to real numbers $\Bbb R$, complex numbers $\Bbb C$, quaternions $\Bbb H$ and octonions $\Bbb O$.  The quaternions are not commutative, and the octonions are additionally not associative.
If the constraint of cancellation is relaxed to not require cancellation of zero divisors generally rather than only the zero vector, then real Clifford algebras exist for every $n=2^m$, with $m$ a nonnegative integer that fit the bill (with every element of the algebra treated as a vector, not only the generating vector space).
If the constraint of bilinearity in the base field is dropped (i.e. we don't require the vector space to be algebra), there will potentially be other possible products.  Though without bilinearity they will be of limited use.

$\dagger$: An algebra over a field is essentially a vector space equipped with a product that is bilinear in the underlying field.  A division algebra is simply an algebra which admits division. 
