Sure (even with the distinctness requirement that wasn't originally there). n1=1, n2=2, ..., and nm = whatever's left. This will only work if n is at least 1+2+

Sure (even with the distinctness requirement that wasn't originally there). n1=1, n2=2, ..., and nm = whatever's left. This will only work if n is at least 1+2+...+m = m(m+1)/2, but if that isn't true then there's no solution.