In fact, a field $k$ can be ordered iff $-1$ is not a sum of squares.
If $-1$ is a sum of squares in an ordered field, then $-1$ is positive whence $1$ is negative; but $1$ is a square and so positive. It is a contradiction.
Let us suppose that $-1$ is not a sum of squares. Then set of field $K \supset k$ in which $-1$ is not a sum of squares is inductive; by Zorn's lemma, there exists a maximal field $\tilde{k}$ containing $k$ in which $-1$ is not a sum of squares.
Let $\gamma \in \tilde{k}$ not be a square. Then $\tilde{k}\subsetneq \tilde{k}(\sqrt{\gamma})$ where $\tilde{k}(\sqrt{\gamma})$ is a rupture field of $X^2-\gamma$. By definition of $\tilde{k}$, there exist $a_i$ and $b_i$ such that $-1= \sum\limits_{i=1}^n (a_i+ \sqrt{\gamma}b_i)= \sum\limits_{i=1}^n (a_i^2+ \gamma b_i^2) + 2 \sqrt{\gamma}\sum\limits_{i=1}^n a_ib_i$. Because $\sqrt{\gamma} \notin \tilde{k}$, $\sum\limits_{i=1}^na_ib_i=0$ and because $-1$ is not a sum of squares in $\tilde{k}$, $\gamma$ is not a sum of squares in $\tilde{k}$.
So, in $\tilde{k}$, a sum of squares is a square. Because $-1$ is not a sum of squares in $\tilde{k}$, $\sum\limits_{i=1}^n b_i^2 \neq 0$. Then we deduce that $\displaystyle - \gamma = \frac{1^2+ \sum\limits_{i=1}^n a_i^2}{\sum\limits_{i=1}^n b_i^2}$ is a square. So, for all $\gamma \in \tilde{k}$, $\gamma$ or $-\gamma$ is a square. Finally, define on $k$ the following order: $y \leq x$ iff $x-y$ is a square in $\tilde{k}$ and prove that $(k,\leq)$ is an ordered field.
For example, $\mathbb{C}$ and $\mathbb{F}_{p^n}$ cannot be ordered because $-1=i^2$ in $\mathbb{C}$ and $-1= \underset{p^n-1}{\underbrace{1^2+...+1^2}}$ in $\mathbb{F}_{p^n}$.
Reference: Algebra, van der Waerden.