It can be proven that (try to prove it!) if some $$$k$$$ works, then $$$k=\displaystyle \left \lfloor \frac{n-1}{2} \right \rfloor$$$ has to work. This means that there is always an answer using $$$n$$$ or $$$n-1$$$ elements, depending on parity.

Actually, there is no problem in repeating $$$1$$$, $$$2$$$, or any other power of two. This is because it can be proven that (try to prove it!) those sums are not optimal.

This solution uses the fact that $$$m\le 10^6$$$, other solutions do not use this, and work for much larger values of $$$m$$$ (like $$$m\leq 10^{18}$$$, but taking the answer modulo some big prime number). Try to solve the problem with this new constraint!