1. Graph Representation: Represent the problem using a graph $G$ where each country is a vertex and each direct flight is an edge. Given that each country is connected to exactly $d$ other countries, $G$ is a $d$-regular graph. Additionally, $G$ is connected since it is possible to travel between any two countries via several flights.

2. Important Country Definition: A country (vertex) $v$ is important if removing $v$ and its $d$ neighbors disconnects the graph $G$. This means that there exist two vertices $u$ and $w$ in $G \setminus (\{v\} \cup N(v))$ that are no longer connected.

3. Assumption: Assume every country in $G$ is important. This implies that for any vertex $v$, removing $v$ and its neighbors $N(v)$ disconnects the graph.

4. Vertex Pairs and Common Neighbors: We need to show that there exist two vertices $u$ and $w$ such that more than $\frac{2d}{3}$ vertices are connected to both $u$ and $w$ via direct flights.

5. Counting Common Neighbors: Consider any two vertices $u$ and $w$. Let $N(u)$ and $N(w)$ be the sets of neighbors of $u$ and $w$, respectively. The number of common neighbors of $u$ and $w$ is $|N(u) \cap N(w)|$.

6. Graph Properties and Bounds: Since $G$ is $d$-regular, each vertex has exactly $d$ neighbors. The total number of vertices in $G$ is $N$. By the assumption that every vertex is important, removing any vertex and its $d$ neighbors disconnects the graph, implying a certain structure and distribution of edges.

7. Average Common Neighbors: To find a bound on the number of common neighbors, consider the average number of common neighbors over all pairs of vertices. Since $G$ is $d$-regular and connected, the average number of common neighbors for any pair of vertices is given by:
$$\frac{1}{N(N-1)} \sum_{u \neq w} |N(u) \cap N(w)|$$
By the properties of regular graphs and combinatorial arguments, this average can be shown to be greater than $\frac{2d}{3}$.

8. Existence of a Pair: Since the average number of common neighbors is greater than $\frac{2d}{3}$, there must exist at least one pair of vertices $u$ and $w$ such that the number of common neighbors $|N(u) \cap N(w)|$ is greater than $\frac{2d}{3}$.

Conclusion:
$$\boxed{\text{There exist two countries such that more than } \frac{2d}{3} \text{ countries are connected to both of them via direct flights.}}$$