1. A country has $n(n \geqslant 2)$ cities, and there is a direct two-way flight route between every two cities. The aviation authority authorizes some airlines to use these routes, and the following conditions are met:
(1) Each route is used by exactly one airline;
(2) For each airline, passengers can use the routes of this airline to travel from one city to all other cities.

Find the maximum number of airlines that the aviation authority can authorize.
1. Let the number of authorized airlines be $m$.

Below is the proof: $m \leqslant\left\lfloor\frac{n}{2}\right\rfloor$.
Let the set of airlines be $\{1,2, \cdots, m\}$. For each $j \in\{1,2, \cdots, m\}$, the routes used by airline $j$ are numbered $j$.

By condition (1), each route has exactly one number; by condition (2), passengers can use the routes of airline $k$ to travel from one city to all other cities, so the routes numbered $j$ must be at least $n-1$ (otherwise, if the routes numbered $j$ are at most $n-2$, there would be a city that passengers cannot reach).
Since there are $\frac{n(n-1)}{2}$ routes in total, we have:
$$
\begin{array}{l}
m(n-1) \leqslant \frac{n(n-1)}{2} \\
\Rightarrow m \leqslant\left\lfloor\frac{n}{2}\right\rfloor .
\end{array}
$$

Next, we use mathematical induction on $n$ to prove that the aviation authority can authorize $m=\left\lfloor\frac{n}{2}\right\rfloor$ airlines.

If $n=2$, there are two cities and one airline is authorized, the conclusion holds.

Assume that for all positive integers $l(2 \leqslant l \leqslant n$, $n \geqslant 2)$, the conclusion holds.
We need to prove that for $n+1$, the conclusion also holds.
If $n=2 k\left(k \in \mathbf{Z}_{+}\right)$, consider city $B$ and $2 k$ cities $A_{1}, A_{2}, \cdots, A_{2 k}$ (as shown in Figure 1).

By the induction hypothesis, the routes between cities $A_{1}, A_{2}, \cdots, A_{2 k}$ are authorized to $k$ airlines. Authorize the routes between city $B$ and $A_{i}$ $(1 \leqslant i \leqslant k)$ to airline $i$, and other routes can be authorized arbitrarily, then passengers can use any of the $k$ authorized airlines to travel from cities $A_{1}, A_{2}, \cdots, A_{2 k}$ to $B$.

Therefore, for $n+1=2 k+1$, the aviation authority can authorize $k$ airlines.

If $n=2 k+1\left(k \in \mathbf{Z}_{+}\right)$, consider cities $B, C$ and $2 k$ cities $A_{1}, A_{2}, \cdots, A_{2 k}$ (as shown in Figure 2).

By the induction hypothesis, the routes between cities $A_{1}, A_{2}, \cdots, A_{2 k}$ are authorized to $k$ airlines. For each $i$ $(1 \leqslant i \leqslant k)$, authorize the routes between city $B$ and $A_{2 i}$ to airline $i$, and the routes between city $C$ and $A_{2 i-1}$ to airline $i$. Authorize the routes between $B$ and $C$, $B$ and $A_{2 i-1}$, and $C$ and $A_{2 i}$ to a new airline $k+1$, then passengers can use any of the $k$ old authorized airlines to travel from $A_{1}, A_{2}, \cdots, A_{2 k}$ to $B$ and $C$, and can also use the new airline $k+1$ to travel from $B$ and $C$ to $A_{1}, A_{2}, \cdots, A_{2 k}$. Thus, for $n+1=2 k+2$, the aviation authority can authorize $k+1$ airlines.
Therefore, for all positive integers $n \geqslant 2$, the conclusion holds.
In summary, $m_{\max }=\left\lfloor\frac{n}{2}\right\rfloor$.