I did last year IMO as a mock test xd so I knew it was an imo problem

alpha any even integer works, because 2(1+2+...+n)=n(n+1). We claim these are the only solutions.

If alpha is not an even integer, we can add any even integer to alpha without changing whether the condition holds (due to the same identity), so we may assume that -1<alpha<=1.

Case 1: alpha=1. This fails because 1+2=3 is not a multiple of 2.

Case 2: 0<alpha<1. Let k>1 be the smallest integer such that k(alpha)>=1; note that k(alpha)=(k-1)(alpha)+alpha<1+1=2. Hence taking n=k, the sum is 0+0+...+0+1=1, which is not a multiple of n.

Case 3: -1<alpha<0. Let k>1 be the smallest integer such that k(alpha)<-1; note that k(alpha)=(k-1)(alpha)+alpha>(-1)+(-1)=-2. Hence taking n=k, the sum is (-1)+(-1)+...+(-1)+(-2)=-n-1, which is not a multiple of n.

Hence the only alpha which work are the even integers.