What can be said about the rate of growth of $f(n)$, defined by $$f(n) = \min_{|V(G)|=n} \left[ \chi(G) + \chi(\bar{G}) \right],$$ where the minimum is taken over all graphs $G$ on $n$ vertices.
Two observations.
(1) Either $G$ or $\bar{G}$ contains a clique on roughly $\log{n}$ vertices by Ramsey theory, so $f(n) \ge c_1 \log n $ for some constant $c_1 > 0$.
(2) If $G = G(n,1/2)$ is a random graph, then $\chi(G) \approx \chi(\bar{G}) \approx n / \log{n}$ almost surely, so we also have $f(n) \le c_2 \, n/ \log n$ for some constant $c_2 > 0$.
These bounds seem hopelessly far apart.
Can we improve on the bounds $$ c_1 \log n \le f(n) \le c_2 \, n / \log n$$ for all sufficiently large $n$?
(Answering for posterity, don't select this as best.)
First, extend @ccc's proof to show that if $pq\geq n$ then $f(n)\leq p+q$. You do this first by taking a graph on $n$ nodes consisting of (at most) $p$ cliques of at most $q$ nodes each, and noting that $\chi(G)\leq p$ and $\chi(\bar G)\leq q$.
Now, as shown by ccc in coments above, $$\lceil 2\sqrt{n} \rceil \leq f(n) \leq 2 \lceil \sqrt{n}\rceil$$
This reduces to at most two values.
In general, if $\lceil 2x \rceil < 2\lceil x \rceil$, then the fractional part of $x$ is in $(0,\frac{1}{2})$. So we only need to consider the cases where the fractional part of $\sqrt n$ is in this range, that is, $\sqrt n =p + \delta$ where $p$ is an integer and $0<\delta<\frac{1}{2}$.
Claim: $p(p+1)\geq n$.
We know that $p(p+1)+\frac{1}{4}=(p+\frac{1}{2})^2>(p+\delta)^2=n$. Since $n$ and $p(p+1)$ are integers, this means $p(p+1)\geq n$.
Therefore, $f(n)\leq 2p+1 = \lceil 2\sqrt{n}\rceil$
So we know that $f(n)$ is always $\lceil 2\sqrt{n}\rceil$