Let $B = \{b_1, b_2, \ldots, b_{|B|} \}$ where $b_1 < b_2 < \ldots < b_{|B|}$ and $|B| \ge 3$. Claim: $B$ satisfies property $T$ iff $b_1 + b_2 > b_{|B|}$ Proof: If $b_1 + b_2 > b_{|B|}$, then for any $1 \le i < j < k \le |B|$, we have $b_i + b_j > b_1 + b_2 > b_{|B|} \ge b_k$, so $T$ is satisfied. If $b_1 + b_2 \le b_{|B|}$, then $T$ is not satisfied trivially. We show that if $B$ satisfies property $T$, then $|B| \le 1009$. If $|B| \ge 1010$ and satisfies $T$, then $b_1 + b_2 \le (b_{|B|} - 1009) + (b_{|B|} - 1008) = 2b_{|B|} - 2017$. However $b_{|B|} < b_1 + b_2$, so by transitivity, this implies $b_{|B|} < 2b_{|B|} - 2017 \Rightarrow b_{|B|} > 2017$, a contradiction. We can attain $|B| = 1009$ by setting $B = \{1009, 1010, \ldots, 2017\}$, so $1009$ is maximal.