Minimum cardinal of a set of sums Let $E$ be a finite set of $n\ge2$ real numbers and let $S=\{a+b\mid a,b\in E\}$.
What is the best minimum number of elements of $S$ ?

I have tried finding a trend with low $n$s, but didn't find anything.
One can easily see that the maximum number of elements of $S$ is $n(n+1)/2$. For $n=2,E=\{a,b\}$, $S$ has exactly three elements. For $n=3,E=\{a,b,c\}$, by setting $a=0,b=-c$ one can get a set of only $5$ elements, which is the minimum. For higher $n$s, it becomes clearly harder to compute the minimum number of elements...

The maximum is indeed
$$\max|S|={n \choose 2}=\frac{n(n+1)}{2}$$
The minimum is achieved with an arithmetic sequence:
$$E=\{k: \frac{-n}{2}\leq k\leq \frac{n}{2},k\in N\}$$
An arithmetic sequence have the property that:
$$a_n=\frac{a_{n+m}+a_{n-m}}{2}$$
Now lets have a look at $S$
$$S=\{k : -n<k<n\}$$
So we get $2n-1$ items in total,
$$\min|S|=2n-1$$
Proof of minimalism
$$|E|\geq 2 \rightarrow \max(E)\neq \min(E)$$
We'll denote $e_1=\min(E)$ and $e_n=\max(E)$
Define:
$$A=\{e_i+e_1:e_i\in E-\{e_n\}\}$$
$$B=\{e_i+e_n:e_i\in E-\{e_1\}\}$$
Obviously:
$$A,B,\{e_1+e_n\}\subset S$$
lets assume that $a_i=a_j$ and $i\neq j$
$$e_1+e_i=e_1+e_j \rightarrow e_i=e_j \rightarrow i=j$$
lets assume that $a_i=b_j$
$$e_1+e_i=e_n+e_j \rightarrow e_n-e_1=e_i-e_j$$
but
$$e_1<e_i,e_j<e_n$$
by definition.
Therefore the size of $|S|$ is at least
$$|A|+|B|+|\{e_1+e_n\}|=(n-1)+(n-1)+1=2n-1$$
Examples:
$$E_1=\{-1,0,1\}$$
$$S_1=\{-2,-1,0,1,2\}$$
