Let $V$ be a vector space over $\mathbb{R}$. A complex structure on $V$ is a linear map $J:V\to V$ such that $J^2=-\text{Id}$.
Remark 1: if $V$ has a complex structure, then $V$ is even dimensional. In fact, $$ J^2 = -\text{Id} \implies (\det J)^2 = (-1)^{\dim V}, $$ which has no solutions over $\mathbb{R}$ if $\dim V$ is odd.
Remark 2: let $W \subset V$ be a subspace with $\dim W = \frac{1}{2}\dim V$. Then $V = W\oplus JW$. Notice that, due to the dimension of $W$, we only have to check that $w$ and $Jw$ are linearly independent for any non-zero $w\in W$.
Indeed, suppose that $ w = c\cdot Jw$ for some $c\in\mathbb{R}$. Then, applying $J$ to both sides of the equality, $J(w) = -c\cdot w$. Combining the equations, we find that $(1+c^2) w = 0$. Since $w\neq 0$, we must have $c^2 + 1 = 0$, which has no solutions over $\mathbb{R}$.
Analogously, let $V$ be a vector space over $\mathbb{C}$. A quaternionic structure on $V$ is an antilinear map $J:V\to V$ such that $J^2=-\text{Id}$.
I believe Remark 1 and 2 above are reproduced in this case. However, the proofs presented above are not analogously extended to this case due to the following reasons.
Remark 1: the proof fails since $(\det J)^2 = (-1)^{\dim V}$ has $\pm i$ as solutions over $\mathbb{C}$.
Remark 2: The proof fails since $c^2 + 1 = 0$ has $\pm i$ as solutions over $\mathbb{C}$.
Did I make a mistake?
Indeed, the problem was treating $J$ as a linear rather than antilinear (= linear conjugate) map. Remember that this means $J(av+bw) = \bar{a} J(v) + \bar{b}J(w)$.