Let us say that the airport $B$ is accessible from the airport $A$ if one can reach $B$ starting from $A$ by making connections of direct flights.

First, we will show that the answer we seek is no more than $1004^2$.

In the sequel until we say otherwise, we will assume that we are in the situation where one of the direct flights was canceled from the original set-up in which the operation of direct flights satisfied both of the conditions (1) and (2) of the problem. Then, we see that there is a positive integer $n$ such that we can divide the set of the $2010$ airports into subsets $V_1, V_2, \dots, V_n$ in such a way that the following conditions (a), (b), (c) are satisfied.

(a) None of the subsets $V_i$ ($1 \le i \le n$) is empty, and every airport belongs to one and only one of the $V_i$'s.
(b) For each $i$ ($1 \le i \le n$), for any pair of different airports $A$ and $B$ belonging to $V_i$, $B$ is accessible from $A$.
(c) For any pair of different $i, j$ ($1 \le i, j \le n$), there are different airports $A$ and $B$ belonging to $V_i$ or $V_j$ such that $B$ is not accessible from $A$.

For different groups $X$ and $Y$ of airports, we say that $Y$ is accessible from $X$ if the following condition is satisfied:
There exist airports $A$ and $B$ such that $A$ belongs to $X$, $B$ belongs to $Y$ and $B$ is accessible from $A$.

We also say that a group $X$ of airports is a startable group if there exists a group $Y$ of airports such that $Y$ is accessible from $X$. If there is no such group $Y$, then $X$ is called a non-startable group. $X$ is called an accessible group if there exists a group $Y$ such that $X$ is accessible from $Y$. If there is no such group $Y$, then $X$ is called a non-accessible group.

Let us first show that there must exist a non-startable group and a non-accessible group. So, suppose on the contrary every group of airports is startable. Then, there exists a sequence $U_1, U_2, \dots$ of airports such that for every $i \ge 1$, $U_{i+1}$ is accessible from $U_i$. Since the number $n$ of groups is finite, there exist integers $j < k$ for which $U_j = U_k$. Since $U_j$ and $U_{j+1}$ are different groups, we must have $j < j+1 < k$. But since each group of airports satisfies the condition (b), every airport in the group $U_{j+1}$ is accessible from every airport in the group $U_j$, and every airport in the group $U_k = U_j$ is accessible from every airport in the group $U_{j+1}$. But this contradicts the condition (c). Thus we have shown that there must exist a non-startable group. Similarly, we can show that there must exist a non-accessible group.

Now, let $X$ be one of the non-startable groups, and $Y$ be one of the non-accessible groups. Suppose that the direct flight canceled was from airport $A$ to airport $B$. Since the condition (1) of the problem was satisfied before the flight was canceled, $A$ must belong to the group $X$ and $B$ to the group $Y$. If $X = Y$, then by the condition (b) it would be possible to reach $B$ from $A$ without using the direct flight, and this would contradict the fact that the condition (2) of the problem was satisfied before the cancelation. Therefore, we must have $X \neq Y$. We also see that $A$ was accessible from $B$ before the cancelation, and there was a way to reach $B$ from $A$ without using the direct flight from $A$ to $B$ before the cancelation and hence $B$ is accessible from $A$ even after the cancelation. So, let us suppose that $B = C_1, C_2, \dots, C_p = A$ is the route of connecting direct flights starting from $B$ and reaching $A$. Now, choose airports $A'$ and $B'$ in such a way that the following 2 conditions (i) and (ii) are satisfied:

(i) $A'$ belongs to the group $X$ and $B'$ belongs to the group $Y$.
(ii) There exists a way to reach $A'$ from $B'$ using neither a direct flight between 2 airports belonging to $X$ nor a direct flight between 2 airports belonging to $Y$.

For example, among the $C_l$'s belonging to $Y$, let $C_{l'}$ be the one corresponding to the largest index and define $B' = C_{l'}$, and define $A'$ to be the $C_m$ corresponding to the smallest index among $C_m$ ($m > l'$) belonging to $X$.

Let us now show by opening a direct flight we can attain a situation where both the conditions (1) and (2) are satisfied. Let us denote by $A_1, A_2, \dots, A_s$ and $B_1, B_2, \dots, B_t$ the airports belonging to the group $X$ and $Y$, respectively. In order to satisfy the condition (1), we see that we have to open a direct flight starting from an airport in the group $X$ and arriving at an airport belonging to the group $Y$. Therefore, the number of ways to open a direct flight to achieve our goal is at most $st$. Furthermore, we have the condition $s + t \le 2010$. Thus, we see that if

$s=1$ or $t=1$ is satisfied, the number of ways is less than or equal to $2009$. We assume in the sequel that both $s$ and $t$ are greater than or equal to $2$.

Since $X$ satisfies the condition (b), there exists an airport $A''$ in $X$ such that there exists a direct flight from $A''$ to $A'$. Similarly, there exists an airport $B''$ in $Y$ such that there exists a direct flight from $B'$ to $B''$. Suppose opening a direct flight from $A_i$ to $B_j$ we can get the situation where both of the conditions (1) and (2) are satisfied. We can show that $A_i \neq A''$. Suppose on the contrary we have $A_i = A''$. Then after opening the direct flight $A_i \to B_j$, we get a direct flight from $A''$ to $B_j$, and by the condition (b) we can reach $B'$ from $B_j$ by using several direct flights between 2 airports belonging to $Y$, and by the condition (ii) we can reach $A'$ from $B'$ without using any direct flight between 2 airports belonging to $X$. Thus we can reach from $A''$ to $A'$ without using the direct flight from $A''$ to $A'$. If we then cancel the direct flight from $A''$ to $A'$ the condition (1) remains valid, so we get a contradiction to the fact that the condition (2) must be satisfied after the direct flight $A_i \to B_j$ is opened. Therefore, we must have $A_i \neq A''$. By a similar argument we can also show that $B_j \neq B''$.

We thus see that the number of ways of opening a direct flight so as to have the conditions (1) and (2) satisfied is at most $(s-1)(t-1)$. Since $s+t \le 2010$ is also satisfied, we see that the desired number is at most $1004^2$.

Let us finally show that there is an example where there are $1004^2$ ways of opening a direct flight to have the conditions (1) and (2) satisfied.

Suppose for $2010$ airports $A_1, A_2, \dots, A_{1005}, B_1, B_2, \dots, B_{1005}$ there are direct flights from $A_i$ to $A_{i+1}$, from $B_{i+1}$ to $B_i$ for each $i$ ($1 \le i \le 1004$), direct flights from $A_{1005}$ to $A_1$ and $B_1$ to $B_{1005}$, and direct flights from $A_4$ to $B_4$ and from $B_2$ to $A_2$. We can easily check that the conditions (1) and (2) are satisfied with this situation. Now suppose we cancel the direct flight from $A_4$ to $B_4$. Then by opening any direct flight from $A_i$ to $B_j$ where $2 \le i \le 1005$ and $2 \le j \le 1005$ we can get the situation where the conditions (1) and (2) will be satisfied. Thus there are at least $1004^2$ ways of opening a direct flight to have both of the conditions (1) and (2) are satisfied.

We have thus shown that the desired answer to the problem is $1004^2 = 1008016$.