China Girls' Mathematical Olympiad

By the definition of $A$, we have $k(k+1)\ge 19$, implying $k\ge 4$.

If $k=4$, we have $k(k+1)=20$. We may assume that $a_1<a_2<a_3<a_4$. Then, $a_4\ge 10$. (1) If $a_4=10$, then $a_3=9$, and we have $a_2=8$ or $7$. If $a_2=8$, then $20,10-9=1,9-8=1$, impossible. If $a_2=7$, then $a_1=6$ or $5$. Since $20,10-9=1,7-6=1$ or $20,9-7=2,7-5=2$, impossible. (2) If $a_4=11$, then $a_3=8$, and we have $a_2=7$ and $a_1=6$, impossible. (3) If $a_4=12$, then $a_3=7$, and we have $a_2=6$ and $a_1=5$, impossible. (4) If $a_4=13$, then $a_3=6$, $a_2=5$, $a_1=4$, impossible. (5) If $a_4=14$, then $a_3=5$, $a_2=4$, impossible. (6) If $a_4=15$, then $a_3=4$, $a_2=3$, $a_1=2$, impossible. (7) If $a_4=16$, then $a_3=3$, $a_2=2$, $a_1=1$, impossible. (8) If $a_4\ge 17$, impossible. So $k\ge 5$. Let $A=\{1,3,5,9,16\}$, then $A$ satisfies the conditions of the problem. Therefore $k_{\min }=5$.