What kind of thing is the set of all vectors in $\mathbb R^n$ with non-negative components? The set of vectors in $\mathbb R^n$ with components greater than or equal to zero is not a vector space. However, it shares many properties with a vector space.
It has more structure than just a set. What kind of algebraic thing is it?
 A: It is what you call a convex cone. A cone in a vector space is a set $E$ such that $x \in E, t\geq 0$ implies $tx \in E$. A convex set is a set $E$ such that whenever $x,y \in E$ the entire line segment from $x$ to $y$ (namely $\{ct+(1-c)y: 0\leq c \leq 1\}$) is contained in $E$. A convex cone is  a set which is both a convex set and a cone. Such a set is automatically closed under addition. 
A: A set $G$ with an associative operation $+$ and an element $0\in G$ such that $0+g=g+0=g$ for all $g\in G$ is called a monoid. In that sense, $\mathbf R_{\geq 0}$ and $\mathbf R_{\geq 0}^n$ are (commutative) monoids. One can talk of modules also over monoids. Then $\mathbf R_{\geq 0}^n$ is a free $\mathbf R_{\geq 0}$-monoid.
A: The structure that jumped to mind for me is this. 
Since $\mathbb R_{\geq 0}$ is a semifield, I would have suggested that $\mathbb R_{\geq 0}^n$ is a semimodule over a semifield, which should probably also be called a semi-vector-space.
This is more general than the convex cone idea (it doesn't rely on a notion of order, and semimodules can be used for any semiring) and it is more specific than the idea of calling it a monoid (it's an abelian monoid: every semimodule has an underlying abelian monoid just like a module has an underlying abelian group.)
