Taiwan IMO Selection Camp, 2021

The maximum is $\frac{n}{2}$. This is achieved when exactly half of $a_i$ and exactly half of $b_j$ are $1$, and the others are $0$.

To show that this is the maximum possible, WLOG assume that $a_1, \dots, a_s$ and $b_1, \dots, b_t$ are nonzero, and the rest are zero. Then we have $a_1 + \cdots + a_s = b_1 + \cdots + b_t = n$ and
$$\min(A \cup B) \le \frac{1}{\max(s, t)} \sum_{i=1}^{s} \sum_{j=1}^{t} \frac{a_i b_j}{a_i b_j + 1}. \quad (*)$$
Let $k = st$ and $x_{(i-1)t+j} = a_i b_j$ for all $i = 1, \dots, s$ and $j = 1, \dots, t$. Then $x_1, \dots, x_k > 0$ and $x_1 + \cdots + x_k = (a_1 + \cdots + a_s)(b_1 + \cdots + b_t) = n^2$. Moreover, we have $\max(s, t) \ge \sqrt{k}$.

Therefore
$$\min(A \cup B) \le \frac{1}{\sqrt{k}} \sum_{i=1}^{k} \frac{x_i}{x_i + 1}. \quad (**)$$
Note that the function $f(x) = \frac{x}{x+1}$ is concave for $x > -1$. Therefore
$$\sum_{i=1}^{k} \frac{x_i}{x_i + 1} \le k \cdot \frac{\frac{n^2}{k}}{\frac{n^2}{k} + 1} = \frac{k n^2}{n^2 + k}.$$
As a consequence,
$$\min(A \cup B) \le \frac{\sqrt{k} n^2}{n^2 + k} \le \frac{n}{2} \quad (***)$$
where the last inequality follows from AM-GM.