A quaternion structure on a real vector space $V$ is a pair of operators $J,K \in \mathcal{L}(V)$ such that $$J^2 = K^2 = -I, \; \; JK = -KJ$$ where $-I$ is the negative identity matrix.
This makes $V$ into a left module over the quaternions and in particular $\mathrm{dim}(V)$ is a multiple of $4$. I wonder if there is a "direct" way to see this, similar to the proof that $\mathrm{dim}(V)$ is even because $\mathrm{det}(J)^2 = (-1)^{\mathrm{dim}\, V}$ is positive.
I have just noticed this old post, but here's a possible "direct" way to see the divisibility by 4:
we know that the existence of J already implies that $\mbox{dim}V$ is even, say $\mbox{dim}V=2n$, and that this gives $V$ the structure of complex vector space with $J$ acting as $i$.
Fix then a complex basis of $V$, say $\{e_1,...,e_n\}$, and consider the real basis of $V$ given by $\{e_1,..., e_n, ie_1,..., ie_n\}$. In such a basis the matrix of $J$ is just the block matrix given by $$J=\begin{pmatrix} 0 & -Id \\ Id & 0 \end{pmatrix}$$ From $JK=-KJ$ it follows easily that the matrix of $K$ in such a basis has to be of the form $$K=\begin{pmatrix} X & Y \\ Y & -X \end{pmatrix}$$ for some real $n\times n$ matrices $X,Y$.
But then $$\begin{pmatrix} X^2+Y^2 & 0 \\ 0 & X^2+Y^2 \end{pmatrix}=K^2=\begin{pmatrix} -Id & 0 \\ 0 & -Id \end{pmatrix}$$ implies that $0\leq \mbox{det}(X^2+Y^2)=(-1)^n$, and so that $n$ has to be even, i.e. that $\mbox{dim}V=2n$ has to be divisible by 4.
P.S. The claimed inequality just comes from $$\mbox{det}(X^2+Y^2)=\mbox{det}((X+iY)(X-iY))=\mbox{det}(X+iY)\overline{\mbox{det}(X+iY)}\geq 0$$