Here is the problem: 
Here is my approach: I attempted the following: I constructed a graph on $[N]=\{1,\cdots,N\}$, adding an edge between $a,b$ in $[N]$ iff $|a-b| \in A.$ Then each $(x,y,x+y) \in A^3$ correspond to triangles of the form $\{a, a+x, a+x+y\}$ or $\{a, a+y, a+x+y\}$. This has $< \delta n^3$ triangles. Thus, by Graph Regularity Lemma, I can remove $\epsilon n^2$ triangles such that this graph is triangular free.
I also tried to do this on a tripartite graph on $X \times Y \times Z$ where each part is isomorphic to $\mathbb{Z}/(2N+1)\mathbb{Z}$ and draw an edge between $a \in X, b \in Y \iff b-a \in A$. Same for X,Z and Y,Z. Thus, by Graph Regularity Lemma, I can remove $\epsilon n^2$ edges such that this graph is triangular free. (This approach makes counting edges far easier)
The hurdle that neither approach can pass is that removing $\epsilon n^2$ edges doesn't correspond to removing $\epsilon n$ elements from $A$. A few possible ideas for getting over the hurdle that I have not gotten to work yet:
I can make my partition of the graph a refinement of a fixed partition and still have bounded # of parts (say refinement of $\{X,Y,Z\}$); energy increases by at least a fixed amt every step, so I get an epsilon-regular partition. Then I can still apply the triangular removal lemma.
(Say I have $> \delta/2 n^3$ triangles in the beginning) The lemma says that if I expect an edge at random, I get rid of $O(n)$ triangles (the expected number is 3*#triangles / #edges). Therefore, I can show that after getting rid of $\alpha n$ edges for some suitably small $\alpha$ (based on $\delta$; if $\delta$ is arbitrarily small then this \alpha can be arbitrarily small), there are $o(n^3)$ triangles. However, the efficiency also gets terrible when #triangle/n^3 decreases (I do not know how to handle it if this is $o(1)$).
I tried to apply Roth for 3-sum instead of 3-AP, which also didn't really work since large counterexamples exist for $\mathbb{Z}/m\mathbb{Z}$