>>457
Pre-School level proof.

Let $m:=\frac{n(n-1)}{2}$, $M:=\{1,\ldots,m\}$ and $D:=\{a_i-a_j: a_i>a_j, 1\leq i,j\leq n\}=\{d_1,\ldots,d_m\}$ be the multiset of differences.
Then let $A:=M\cap D$, $B:=M\backslash A=\{b_1,\ldots,b_t\}$, $C:=D\backslash A=\{c_1,\ldots,c_t\}$. Here $t$ is the number wanted to be larger than $\lceil \frac{n(n-6)}{19} \rceil$.
Consider the generating function $F(z):=z^{a_1}+\cdots+z^{a_n}$, $G(z):=F(z)F(\frac{1}{z})=n+\sum (z^{d_k}+z^{-d_k})$, $H(z):=z^{-m}+z^{-m+1}+\cdots+1+z+\cdots+z^m$. We have $G(z)=n-1+H(z)-\sum (z^{b_l}+z^{-b_l}) + \sum (z^{c_l}+z^{-c_l})$.
Take $|z|=1$, then $G(z)=|F(z)|^2\geq 0$ and $|G(z)-(n-1)-H(z)|\leq 4t$. Write $z$ as $e^{2i\theta}$ then $H(z)=\frac{\sin(2m+1)\theta}{\sin\theta}$. Take $(2m+1)\theta=\frac{3}{2}\pi$, we have $H(z)=-\frac{1}{\sin\theta}<-\frac{n^2-n+1}{\frac{3}{2}\pi}$. Hence $|G(z)-(n-1)-H(z)|\geq \frac{2}{3\pi}(n^2-(1+\frac{3\pi}{2})n)$. So we can get $t\geq \frac{1}{6\pi}n(n-6)$.

Great problem.