To prove that the number of ways to introduce a one-way traffic on each airline such that no city can be returned to (using other one-way airlines) is not divisible by 3, we can use the properties of the chromatic polynomial of a bipartite graph.

1. Restate the problem in graph theory terms:
We are given a bipartite graph $G$ and need to show that the number of acyclic orientations of $G$ is not a multiple of 3. It is known that the number of acyclic orientations of a graph $G$ is given by $|\chi_G(-1)|$, where $\chi_G$ is the chromatic polynomial of $G$.

2. Chromatic polynomial evaluation:
The chromatic polynomial $\chi_G(k)$ of a graph $G$ counts the number of ways to color the vertices of $G$ with $k$ colors such that no two adjacent vertices share the same color. For a bipartite graph $G$, the chromatic polynomial can be evaluated at specific points to derive properties about the graph.

3. **Evaluate $\chi_G(-1)$:**
We need to show that $3 \nmid \chi_G(-1)$. To do this, we use the known result that for a bipartite graph $G$, $\chi_G(-1) \equiv \chi_G(2) \pmod{3}$.

4. **Evaluate $\chi_G(2)$:**
For a bipartite graph $G$, the chromatic polynomial evaluated at 2, $\chi_G(2)$, counts the number of ways to color the graph with 2 colors. Since $G$ is bipartite, it can be colored with 2 colors in exactly $2^c$ ways, where $c$ is the number of connected components of $G$.

5. Modulo 3 calculation:
Since $\chi_G(2) = 2^c$, we need to evaluate $2^c \pmod{3}$. Note that:
$$2^1 \equiv 2 \pmod{3}, \quad 2^2 \equiv 1 \pmod{3}, \quad 2^3 \equiv 2 \pmod{3}, \quad 2^4 \equiv 1 \pmod{3}, \quad \text{and so on.}$$
Therefore, $2^c$ modulo 3 alternates between 2 and 1 depending on whether $c$ is odd or even.

6. Conclusion:
Since $2^c \equiv 1$ or $2 \pmod{3}$, it follows that $2^c$ is never divisible by 3. Hence, $\chi_G(2) \not\equiv 0 \pmod{3}$.

7. Final step:
Since $\chi_G(-1) \equiv \chi_G(2) \pmod{3}$, and $\chi_G(2) \not\equiv 0 \pmod{3}$, it follows that $\chi_G(-1) \not\equiv 0 \pmod{3}$.

$\blacksquare$