Maybe it's $1008$. We just want to find the domination number of a $2017$-tree. For more details, see :http://mathworld.wolfram.com/DominationNumber.html

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smy2012's approach does not work because there is no direct flight from a city to itself.

MustafaKemal wrote: ............ Hi, can you tell me where to find the official solution to Turkey TST?

We'll show that the answer is $1344=2\times \frac{2017-1}{3}$. The example I've is the same with the one used in #4, so I won't write it again here. Now, let's re-written the problem in graph theoretic form: Given connected simple graph $G$ with $3m+1$ where $m\in \mathbb{Z}^+$ vertices, show that we can select $2m$ vertices so that every vertex having at least one selected neighbor. We'll prove by induction on $m\in \mathbb{Z}^+$, the base case is easy. For the inductive step, suppose the result is true when $m\leftarrow t$, we want to prove it's true when $m\leftarrow t+1$: Note that, if there's a cycle in our graph, we can delete one edge in the cycle without deleting the connectivity of the graph. So, this procedure must end when our graph is reduced to one spanning tree. Let the root of our tree is $V$. If the length of path from (one of) the farthest leaf, called $L$, to $V$ is not smaller than $3$, let the path be $V,P_1,P_2,...,P_k,L$ where $k\geq 2$. Select the vertices $P_k$ and $P_{k-1}$. Now our condition have been verified when the vertex is $L$ or $P_k$ or $P_{k-1}$. Consider the graph formed by deleting these three vertices, note that the graph remain connected since $P_{k-1}\neq V$. By induction hypothesis, we can select other $2t$ vertices to get the condition verified for all other vertices. Combine this with our two selected vertices gives us $2t+2=2(t+1)$ vertices, done. If the length of path from any leaf to $V$ is smaller than $3$, we can named the vertices of our graph $O_1,O_2,...,O_a,T_1,T_2,...,T_b,Q_1,Q_2,...,Q_b$, for some non-negative integers $a,b$ that $a+2b=3(t+1)$, so that all edges of our tree are $VO_1,VO_2,...,VO_a,VT_1,VT_2,...,VT_b,T_1Q_1,T_2Q_2,...,T_bQ_b$. In this case, just select $b+1$ vertices $V,T_1,T_2,...,T_b$, not hard to see that the condition is true. Also, $b+1=\frac{2b+2}{2}\leq \frac{3(t+1)+2}{2}=\frac{3t+5}{2}\leq 2(t+1)$, the last inequality is true for all $t\geq 1$. So, we've selected not more than $2(t+1)$ vertices, done.