I'm not 100% sure for this, but you can't subtract this way in ECC.
Think of this. 1*G - 1*G would be equal with 1*G + (-1)*G, right? But, -1 is outside the range. In finite fields, -1 is equal with N-1 which is 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364139.
So, what you really do is 1*G + (N-1)*G which is N*G. Apparently, k*G + (-k)*G is always N*G, but I need this to be confirmed. Anyway, check:
How to subtract two points on an elliptic curve?yes, by -1*G, I mean also (N-1)*G.
I'm sure of my calculations. At least the apparent result is that (-k*G+k*G)= N*G only for (k=1). This could be possible since the addition of pairs (-k*G,k*G) on the curve form parallel lines.
Then, the question is about the link between the sums.
For K=1:
(20860373710434931919338205163613943089612844706565330860427609212248504954008, 28589774168867891354814021154080592739878697919018203946384876578134205873990) = N*G
For K=2:
(49667982834466148699028630885550314015746939561788444090107179747772288677390, 37758011528734597361759657276645959964664126776856991748971785837209648797521) = x*G