There exists a set $A$ on $\{1,...,N\}$ where $ |A|=\frac{2}{\sqrt{3}}\sqrt{N}$$ |A|\geq\frac{2}{\sqrt{3}}\sqrt{N}$ and $E(A)=o(N)$.
Let $B$ be a Sidon set on $\{1,...,n\}$. Let $C = \{3n+1-b | b\in B\}$. Let $A=B\cup C$.
Suppose $a<b\leq c<d$ and $a+d=c+b$, where $a,b,c,d$ are elements of $A$. Now I will analyze the possibilities of $a,b,c,d$.
They can't be all in $B$ or all in $C$, as $B$ and $C$ are Sidon sets.
They can't be 3 in $B$ and 1 in $C$, or vice versa. To see this, note that the sum of any two elements in $B$ is smaller than any element in $C$.
So $a,b\in B$ and $c,d \in C$. Since $b-a=d-c$, it follows that $(3n+1-c)-(3n+1-d)=b-a$. Observe that $(3n+1-c)$, $(3n+1-d)$, $b$ and $a$ are all elements of the Sidon set $B$, so we have $(3n+1-c)=b$ and $(3n+1-d)=a$, i.e. $a+d=c+b=3n+1$.
So it's clear that $r_A(k)$ is larger than $2$ only if $k=3n+1$, where it's $O(\sqrt{n})$, which implies $E(A)=o(n)$.
Let $N$ be the largest element of $A$. $ |A|=\frac{2}{\sqrt{3}}\sqrt{N}$$ |A|\geq\frac{2}{\sqrt{3}}\sqrt{N}$ and $E(A)=o(N)$, as required.