(Just to bump and give a reference) This PNT's problem implies the desired

We will generalize $2016$ by $3n$. Assume contradiction (there doesn't exists $3$ numbers like that) Assume $1 \in A$ and $k \in B$ is the minimum number that not in $A$. We will prove that $c-1 \in A \forall c \in C$. We have $\forall c \in C, c-1 \notin B$ (else we will have $(1,c-1,c)$ that satisfied) Assume exists $c,c-1 \in C$ Now, considering the trio $(c-k,k,c)$ with $c \in C$. Since $k \in B,c \in C$, therefore $c-k \notin A$. Considering $(k-1,c-k,c-1)$ we also have $c-k \notin B$. Therefore, $c-k \in C$. Similarly, by considering $(c-k-1,k,c-1)$ and $(1,c-k-1,c-k)$ we infer $c-k-1 \in C$. Inducting and we have $c-ik,c-ik-1 \in C \forall i \geq 0, i<\dfrac{c-1}{k}$. But obviously there exists $i$ satisfied $0 \leq i <\dfrac{c-1}{k}$ and $c-ik \leq k$, so $c-ik \in A$ or $B$, contradiction. Therefore, $\forall c \in C,c-1 \in A$. If $C=\lbrace c_1,c_2,\cdots , c_n\rbrace$, then $\lbrace c_1-1,c_2-1,\cdots, c_n-1 \rbrace \subset A$ Since $\min C >k>1$, we can see $1 \notin \lbrace c_1-1,c_2-1,\cdots, c_n-1 \rbrace$. So $\lbrace 1,c_1-1,c_2-1,\cdots, c_n-1 \rbrace \subset A$, or $A$ has more than $n$ elements, contradiction.