An alternate Solution:Epshita32 wrote:P7. There are 2000 points on a circle and each point is given a number which is equal to the average of the two numbers which are its nearest neighbors. Show that all the numbers must be equal.

We use extremal principal. Since there a finite number of numbers, there must exists a minimum number. If the minimal number occurs more than once, randomly pick one. Due to minimality, the numbers on the both sides of those numbers would be equal to the minimal number. This way, all the numbers on the circle will be equal to the minimal number and thus we are done.