I will give you an intuitive explanation.

Suppose the array is A = {a1, a2, a3, a4, ...., an}. The only operation allowed is take two elements ai, aj and replace ai by ai-1 and aj by aj+1 or vice versa. Let's take sum S = a1 + a2 + .... + an. We can see that after each such operation, sum remains same because we increase one element whereas decrease the other by 1.

Also notice that if we fix an element a1, we can use it to set rest of the elements to a given value. So, we can make at least (n-1) elements same. Can we achieve n?

Now, since we know that after such set of operations, final sum will be same and to achieve all equal we must have a1 = a2 = .... = an = k. Then k * n = S, which means sum must be divisible by n. This is a necessary condition. To see that it is sufficient too, we use previous knowledge. Lets make all a2, a3, ...., an to 0. Since sum is constant, we will have a1 = S after making rest of the elements 0. Now, we have found previously that using a1, we can make any of the element equal to some value which means we can use a1 to make a2 = k = S / n. Similarly we can make a3 = k = S / n. Finally after making all a2, a3,....an to S/n, a1 will be equal to S — (n-1)S/n = S/n. Thus all of them are equal. This proves (S % n == 0) is sufficient too.