Taiwan IMO Selection Camp

The answer is $k=3$. For example, take
$$A_1 = \{1, 2, 3\} \cup \{3m\mid m \ge 4\}, \quad A_2 = \{4, 5, 6\} \cup \{3m-1\mid m \ge 4\}, \quad A_3 = \{7, 8, 9\} \cup \{3m-2\mid m \ge 4\}.$$
To verify that the above partition satisfies the requirement of the problem, first observe: the situation of the sum of two distinct elements $a, b$ in $A_i$:
(i) $a+b=n \ge 1+12=13$, when $i=1$,
(ii) $a+b=n \ge 4+11=15$, when $i=2$,
(iii) $a+b=n \ge 7+10=17$, when $i=3$.
Therefore, we must find two distinct elements in $A_3$ whose sum is $15(=7+8), 16(=7+9)$.
Now suppose that for some $k \ge 4$, there exist $A_1, A_2, \dots, A_k$ satisfying the requirement of the problem. Clearly, $A_1, A_2, A_3, A_4 \cup \dots \cup A_k$ also satisfies the requirement of the problem, so we may assume $k=4$.
Take $B_i = A_i \cap \{1, 2, \dots, 23\}, i = 1, 2, 3, 4$. For any index $i$ and the 10 numbers 15, 16, ..., 24, each can be represented as a sum of two distinct elements of $B_i$. Therefore, $B_i$ has at least 5 elements. Since $|B_1| + |B_2| + |B_3| + |B_4| = 23$, we have $|B_j| = 5$ for some $j$. Let $B_j = \{x_1, x_2, x_3, x_4, x_5\}$. The sums of two distinct elements in $A_j$ can be 15, 16, ..., 24 and these should be sums of any two elements in $B_j$, that is, $4(x_1 + x_2 + x_3 + x_4 + x_5) = 15 + 16 + \dots + 24 = 195$ and 4 divides 195. This is a contradiction!