Half. We can show by a simple induction that \(F_{2n}=\sum_{1 \leq k \leq n}F_{2k-1}\). Then, we can use all odd-indexed Fibinacci numbers as the \(x_i\) list. How can we prove we can't do better? Maybe a bit of induction. If we catch the set of all numbers used to generate the first \(k\) Fibonacci numbers, the best we can obtain by summing all of them is \(F_1+F_2+F_3+\ldots+F_k=F_{k+1}-1 < F_{k+1}\), then it is impossible to add more than it. Then, we need at least one more if we add the next Fibonacci (maybe \(F_{k+1}\) itself, maybe \(F_{k-1}\) to make a pair). P.S.: there are other 1009-tuples satisfying this problem?

We started by showing the same as above. The rest goes as follow: It follows by the above that all of the numbers $F_1,...,F_{2018}$ can be written as a sum one or more of the $1009$ numbers $F_1,F_3,...,F_{2017}$. Assume that $m$ is minimal. Thus $m\leq 1009$. It can be assumed that $x_1\leq ...\leq x_m$. Since $F_1=1$ can be written as a sum of the numbers $x_1=1$. We will now show by induction that $x_k\leq F_{2k-1}$ for $k=1,...,m$. It is true for $k=1$. Assume that $x_i\leq F_{2i-1}$ for $i=1,...,k-1$. Then $$x_1+...+x_{k-1}\leq F_1 + ... + F_{2k-3}= F_{2k-2}<F_{2k-1}$$If $x_k>F_{2k-1}$ then it is impossible to write $F_{2k-1}$ as a sum of numbers on the board - a contradiction since $2k-1\leq 2017$. Thus $x_k\leq F_{2k-1}$ which completes the indution. If $m<1009$ then $F_{2018}=F_1+...+F_{2017}>F_1+...+F_{2015}\geq x_1+...+x_m$ and $F_{2018}$ cannot be written as a sum of numbers on the board which is a contradiction. Hence the least possible value of $m$ is $1009$.

I don't get it. Why the trivial solution doesn't work?$m=1, x_1=1$ and all of Fibonacci numbers I can express simply as the sum of one's. Please help...