3. On three islands, there are 2019 cities, and some two-way air routes have been opened between them. Each route connects three cities on different islands. Any two cities are on at most one route. Find the maximum number of routes.
3. $673^{2}$.

Assume that there are 673 cities on each of the three islands, denoted as $A_{i}, B_{j}, C_{k}(i, j, k \in\{1,2, \cdots, 673\})$. When the triplet $(i, j, k)$ satisfies 673 । $(i+j+k)$, a flight route connecting cities $A_{i}, B_{j}, C_{k}$ is established. It is easy to see that for each pair of cities $A_{i}, B_{j}$, the third city $C_{k}$ is unique. Similarly, for each pair of cities $B_{j}, C_{k}$, the third city $A_{i}$ is also unique; for each pair of cities $A_{i}, C_{k}$, the third city $B_{j}$ is also unique.
Thus, the number of flight routes opened is $673^{2}$.
Now assume that the number of cities on the three islands are $a, b, c$, without loss of generality, assume $a \leqslant b \leqslant c$. Similarly, denote these cities as $A_{i}, B_{j}, C_{k}(1 \leqslant i \leqslant a, 1 \leqslant j \leqslant b, 1 \leqslant k \leqslant c)$.
According to the problem, the number of flight routes does not exceed $a b$.
By the assumption, we have $a+b \leqslant 1346$.
Thus, $a b \leqslant\left(\frac{a+b}{2}\right)^{2} \leqslant 673^{2}$.
Therefore, the maximum number of flight routes is $673^{2}$.