I was not sure how to even approach this problem, so I started by implementing a brute force solution that found me the answers for all values of n up to 15. The brute force returns that for n=3,4 we can make all numbers equal to 4, for n=5,6,7,8 we can make all numbers equal to 8, and for n=9,10,11,12,13,14,15 we can make all numbers equal to 16. This naturally suggests that we need to target the smallest power of two that is at least n. I did not prove this fact during the contest, but there exists a very simple proof.
Apart from this observation, looking at the way the brute force achieved the goal helped me learn a couple of tricks that are useful for the solution. First of all, when we have a 0, then we can double any number: (0,x)->(x,x)->(0,2x). Second, the solution often makes many smaller powers of two, and then makes them all equal using this trick.
This way we will convert all numbers into powers of two not exceeding 2k. It turns out that for n>=3 at least two of those powers will be equal and less than 2k, which allows us to use the (x,x)->(0,2x) step to get a zero, and then use this zero to keep multiplying all numbers by two until we have only one zero and many 2k, and finally use the (0,x)->(x,x) step to get rid of the zero.
Thanks for reading, and check back next week!