Proposition: If $A\subset\mathbb{N},\ A$ has natural density $d> \frac{1}{2},$ then $\exists\ N\in\mathbb{N}\ $ such that $\ n>N \implies \exists\ a,b\in A\ $ such that $\ a+b=n.$
Proof sketch: Since $d> \frac{1}{2},\ \exists\ N\ $ such that $\ \left\lvert A\cup \{1,2,\ldots, n\} \right\rvert > \frac{1}{2}\ \forall\ n>N.\ $ Consider any $n>N.$ By considering the $\frac{n}{2}\ $ pairs $\ (1,n-1),\ (2,n-2),\ \ldots,\ \left( n/2,n/2 \right),\ $ by the PHP, both members of one of these pairs must be in $A$, that is, $\ x\in A\ $ and $\ n-x \in A,\ $ as desired.
If we replace $d> \frac{1}{2}$ in the above proposition with $\ d = \frac{1}{2},\ $ then the set of even numbers (or the set of odd numbers) are counter-examples to the proposition. Furthermore, requiring there to be at least one even and at least one odd number in $A$ doesn't resolve this: consider $\left(2\mathbb{N}\setminus\{2^n:n\in\mathbb{N}\}\right)\cup\{7\}.$ Then no such $N$ exists.
But what about if, for example, $A$ has natural density $\ d = \frac{1}{2}\ $ and $A\cap(2\mathbb{N}+1)$ has natural density $\ \frac{1}{4}\ ($ in $\mathbb{N}),\ ($ so that $A\cap(2\mathbb{N})$ also has natural density $\ \frac{1}{4}\ )?$ Would the above proposition hold in this case?
A very straightforward counter-example exists: $\ A = (4\mathbb{N})\ \cup (4\mathbb{N}+1).\ $ Then, $A+A = (4\mathbb{N})\ \cup (4\mathbb{N}+2)\cup (4\mathbb{N}+1),\ $ which is not equal to $\mathbb{N},\ $ and has density $\frac{3}{4}.$