The Eccentric Empire

The answer is $2$ .

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First, we will show that $2$ is attainable.
This should be obvious, just imagine a symmetric case, or even more obvious, there are only two cities.

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Next, we will show that if there are (at least) $2$ such cities, they must be adjacent to each other.
With the assumption of being a tree, this proves that there are at most $2$ such cities.

Let $a,b$ has the least eccentricity $e$ .

Notice that $a,b$ cannot be leaves. If it were, that city it links to has less eccentricity.

Assume $u,v$ are the most far away city for $a,b$ respectively.
So $e=\text{distance}(a,u)=\text{distance}(b,v)$
Further assume the contrary that $a,b$ are not adjacent, so $\exists c$ between $a$ and $b$ .
By the definition of eccentricity, the path from $u$ to $v$ is like this:
$u,\cdots,b,\cdots,c,\cdots,a,\cdots,v$
It is now obvious that $c$ has less eccentricity than $a,b$ .

If $u = v$ , Consider the linkage between $a,b,u=v$ .
It must be one of the three cases: $a\cdots b\cdots u$ , $a\cdots u\cdots b$ , $u\cdots a\cdots b$ .
For the middle one, $u$ has less eccentricity than $a,b$ .
For the other two, $a$ and $b$ has different eccentricity.

Contradiction.

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The answer is $2$ .