6. Let $d(a, b)$ denote the distance between cities $a$ and $b$,
$$S_{i}(a)=\{c \mid d(a, c)=i\},$$

i.e., the set of cities whose distance from city $a$ is exactly $i$.
Assume there exists a city $x$ such that the number of elements in the set $D=S_{4}(x)$ is at least 2551.
Let $A=S_{1}(x)$.
If every city in set $D$ can be reached from city $x$ via a subset $A^{\prime}$ of $A$ in 4 flights, then the subset $A^{\prime}$ is called "important." In other words, every city in $D$ is at a distance of 3 from some element in $A^{\prime}$, i.e.,
$D \subseteq \bigcup_{a \in A^{\prime}} S_{3}(a)$.
For example, $A$ itself is important.
Let $A_{0}$ be an important subset of $A$ with the minimum number of elements, and let this number be $m$, i.e., $m=\left|A_{0}\right|$.

From $m(101-m) \leqslant 50 \times 51=2550$, we know there must exist a city $a \in A_{0}$ such that
$\left|S_{3}(a) \cap D\right| \geqslant 102-m$.
Since $\left|S_{3}(a)\right| \leqslant 100$, there are at most
$$100-(102-m)=m-2$$

cities $c$ in $S_{3}(a)$ such that $d(c, x) \leqslant 3$.
Let $T=\left\{c \in S_{3}(a) \mid d(x, c) \leqslant 3\right\}$. Then
$|T| \leqslant m-2$.
To derive a contradiction, we can construct $m-1$ distinct elements in $T$ corresponding to the $m-1$ elements in the set $A_{a}=A_{0} \backslash\{a\}$.
Thus, $|T| \geqslant m-1$, which is a contradiction.
Since $A_{0}$ has the minimum number of elements, for every $y \in A_{a}$, there exists a city $d_{y} \in D$ that can only be reached from $x$ via $y$ in 4 flights.
Therefore, there exist two cities $b_{y}$ and $c_{y}$ such that the flight sequence
$$x-y-b_{y}-c_{y}-d_{y}$$

reaches $d_{y}$ from $x$.
Since this is the shortest flight sequence, we have
$$d\left(x, b_{y}\right)=2, d\left(x, c_{y}\right)=3 \text {. }$$

Next, we prove that for all $y \in A_{a}$, the $2(m-1)$ cities of the form $b_{y}$ and $c_{y}$ are all distinct.

In fact, for any $y, z \in A_{a}$, since the distances from cities $b_{y}$ and $c_{z}$ to $x$ are different, $b_{y}$ and $c_{z}$ must be different.

On the other hand, if for some $y, z (y \neq z) \in A_{a}$, $b_{y}=b_{z}$, then there exists a path of length 4 from $x$ via $y$ to city $d_{z}$, i.e., $x-y-b_{z}-c_{z}-d_{z}$, which contradicts the selection of $d_{z}$.
Similarly, for any $y \neq z$, $c_{y} \neq c_{z}$.
It remains to prove that for every $y \in A_{a}$, one of the cities $b_{y}$ and $c_{y}$ is at a distance of 3 from $a$, and thus belongs to $T$.
Note that $d(a, y) \leqslant 2$ (since there is a path $a-x-y$).
And $d\left(a, d_{y}\right) \geqslant d\left(x, d_{y}\right)-d(x, a)=3$, by the selection of $d_{y}$, we know $d\left(a, d_{y}\right) \neq 3$.
Thus, $d\left(a, d_{y}\right)>3$.
In the sequence $d(a, y), d\left(a, b_{y}\right), d\left(a, c_{y}\right), d\left(a, d_{y}\right)$, the difference between adjacent terms is at most 1.

Since the first term is less than 3 and the last term is greater than 3, one of the middle terms $d\left(a, b_{y}\right)$ or $d\left(a, c_{y}\right)$ must be equal to 3.