Size of sets with complete double Let $[n]$ denote the set $\{0,1,...,n\}$. A subset $S\subseteq [n]$ is said to have complete double if $S+S=[2n]$. Let $m(n)$ be the smallest size of a subset of $[n]$ with complete double. My questions are:

Question 1: has this function been studied before? Any reference? It looks related to Sidon sets, but not quite the same.
Question 2: It is easy to see that $m(n)\geq 2\sqrt{n}-1$. Can we show that $m(n)\approx 2\sqrt{n}$ asymptotically? (see the comments on how to get to $2\sqrt{2n}$).

I found one similar question (for $\mathbb F_2^n$ instead of integers) here but could not find further useful references.
 A: Short answer: Such a set $S$ is known as a restricted additive basis.

As observed in comments, the general term to search for is [finite]
additive basis: a finite set $S \subset \mathbb{N}$ such that $S+S
\supseteq [2n]$.  Another term is postage stamp problem (not to
be confused with the Frobenius postage stamp problem which is
different).
The extra condition that $S \subseteq [n]$ makes this the
restricted postage stamp problem and the solutions are restricted
additive bases. ("Restricted" means that the elements are restricted
to be in $[n]$.)
It is then (almost) equivalent to ask

*

*given $k$, what is the largest $n$ ...

*given $n$, what is the smallest $k$ ...

such that there exists a restricted additive basis of $k$ elements for
the interval $[2n]$.  More about that "almost" at the end of this
answer.  Let us write $n(k)$ for the largest $n$ when $k$ is given,
and $k(n)$ for the smallest $k$ when $n$ is given.
Here the question asks for $k(n)$.  For $n(k)$ this is OEIS
A006638, where currently the largest entry is $a(47)=734$ by
yours truly (2015).  Some translation needed: That "47" counts only the
nonzero elements, so in fact there are 48 elements; that "734"
means the target interval is $[2n]=[734]$, so in our current notation
it becomes $n(48)=367$, and $k(367)=48$.

To the specific questions:
Q1. The OEIS entry gives some pointers to literature.
The oldest reference that I know is H. Rohrbach, "Ein Beitrag zur additiven
Zahlentheorie", Math. Z., 42 (1937), 1-30, which studies both the
unrestricted and the restricted versions.
Symmetric bases have also been studied, see for example S. Mossige,
"Algorithms for computing the h-range of the postage stamp problem",
Math. Comp. 36 (1981), 575–582.
Q2. About asymptotics: No, one cannot achieve $k(n) \sim
2\sqrt{n}$ asymptotically.  Gang Yu, "Upper bounds for finite additive
2-bases", Proc. AMS 137 (2009), 11-18, shows that
$$\limsup_{n \to \infty} \frac{n}{k_r(n)^2} \le 0.419822,$$
where $k_r(n)$ is the smallest size of a restricted basis for $[n]$.
I believe this is currently the best lower bound for $k_r$.
Translated to our notation (with target interval $[2n]$)
this means the factor in front of $\sqrt n$ is asymptotically at least $2.18264$.

And about the "almost": Generally $k(n)$ increases as $n$
increases, but there are some known cases where $k(n)$ suddenly steps
down. One of them was also noticed in the comments above: "What the
heck is going on at n=31?".  Indeed we have $k(31)=14$ but $k(32)=13$;
$k(51)=18$ but $k(52)=17$; and $k(57)=19$ but $k(58)=18$.
See Table 5 in this 2018 paper by me and coauthors, where the
corresponding problem is studied in two dimensions.
A: Let $S\subset [1,N]$ with $S+S=2N$.   I will show that $S$ must have at least $(2.033 +o(1))\sqrt{N}$ elements for large $N$.  The argument can certainly be improved, but I don't know what the right answer should be. 
Assume that $N$ is large and that $|S| =O(\sqrt{N})$ (else we are done of course).  Let $r(n)$ denote the number of ways of writing $n$ as $a+b$ 
with $a$ and $b$ in $S$.  Apart from the cases $a=b$ which occur for at $O(\sqrt{N})$ values, we have $r(n)\ge 2$ for $n\le 2N$ (since $a+b$ and $b+a$ 
will be counted as two different solutions).  As noted in the problem, we then 
have 
$$ 
4N +O(\sqrt{N}) \le \sum_{n\le 2N} r(n) = |S|^2,  
$$ 
which gave the asymptotic bound $|S| \ge 2\sqrt{N}+O(1)$ that we now wish to improve. 
Decompose $S$ into three sets $A$, $B$, $C$, where $A=S\cap[1,N/3)$, $B=S\cap[N/3,2N/3)$, and $C= S\cap[2N/3,N]$. Say $|A|=\alpha\sqrt{N}$, $|B|=\beta\sqrt{N}$ and $|C| =\gamma\sqrt{N}$, so that we know already that $\alpha+\beta+\gamma\ge 2+o(1)$.  
Consider $A+A$.  This must cover all numbers in $[1,N/3]$ and therefore we must have 
$$ 
|A|^2 \ge \sum_{n\le N/3} r(n) \ge 2N/3+ O(\sqrt{N}).
$$ 
Therefore we must have 
$$ 
\alpha \ge \sqrt{2/3} +o(1). \tag{1} 
$$ 
Similarly $C+C$ must cover all elements in $[5N/3,2N]$ which leads to 
$$ 
\gamma \ge \sqrt{2/3}+ o(1). \tag{2}
$$ 
Now consider $B+B$ and $A+C$.  These sums are all in $[2N/3,4N/3]$ and therefore we obtain 
$$ 
\sum_{2N/3 \le n \le 4N/3} r(n) \ge |B|^2 + 2|A| |C| +O(\sqrt{N}) = (\beta^2+2\alpha\gamma+o(1)) N. 
$$ 
Therefore 
$$ 
|S|^2 =\sum_{n\le 2N} r(n) \ge \Big(\sum_{n< 2N/3} +\sum_{2N/3\le n\le 4N/3} +\sum_{4N/3<n \le 2N} \Big) r(n) 
$$
must be at least
$$
\Big(\frac{8}{3} + \beta^2 + 2\alpha \gamma +o(1)\Big) N. \tag{3}
$$
Put $\alpha+\gamma =2\sqrt{2/3}+\delta$, with $\delta>0$.  Since both $\alpha$ and $\gamma$ must be $\ge \sqrt{2/3}+o(1)$ we must have 
$$
\alpha\gamma \ge \sqrt{2/3}(\sqrt{2/3}+\delta)+o(1) = \frac 23 +\sqrt{\frac 23}\delta +o(1).
$$
If $\delta\ge 2-2\sqrt{2/3}$, then using this in (3) we would get 
$$
|S|^2 \ge \Big( 4 +2 \sqrt{2/3} \delta+o(1)\Big) N \ge 4.599N, 
$$ 
which is more than we claimed. 
Suppose then that $\delta \le 2-2\sqrt{2/3}$, so that $\beta \ge 2-(\alpha+\gamma)=2-2\sqrt{2/3}-\delta (>0)$. Using this in (3) we find 
$$ 
|S|^2 \ge \Big( 4+ 2\sqrt{2/3} \delta + (2-2\sqrt{2/3}-\delta)^2 +o(1) \Big) N.
$$
The right side is smallest for $\delta=0$, yielding 
$$
|S|^2 \ge (4+(2-2\sqrt{2/3})^2+o(1))N,
$$
which gives $|S| \ge (2.0333\ldots +o(1))\sqrt{N}$. 
