for q6 i said that each unit up or down changed the bees vertical position by 1/-1 and a unit left or right would change position 1/-1 horizontal and a half up or down. then i said that you need moves left and right to balance so there is 0 horizontal, which means half of the total moves spent displaces vertically. this means that if the total of the sum is a multiple of 3 it should be possible (unless N is 3 because 3 is greater than 2 which is a third of the sum 1+2+3). however it works for even sums from N down, as you can move a set number in one direction and then trace back exactly the same amount of steps. this means if the sum is divisible by 2 then the bee can still do it, so 3 is still valid.
then you use that n+n-1+...+1 is equal to n squared plus n all over 2, and you show for which n it is a multiple of 2 or 3 (if multiple of 2 number is 4k or 4k+3 and if of 3 then number can be 6k, 6k+2(exclude 2), 6k+3, 6k+5)
im not sure if theres any other possible values and my actual written solution was quite bad so i can imagine having missed out something important, hopefully this years boundaries are much lower