There can't be two consecutive positive integers coz if there is two consecutive +ve numbers then the next number will again positive as it is greater than the sum of previous two.And again the next number will positive So all the number will positive.But now look at the maximum number among these 2018 numbers.As the next number in clock wise order will less than it( by maximality of that number)and is greater than sum of previous 2 so the previous number of the max number must be negative.We get a contradiction of the fact that all the number is positive.
So among any 2 consecutive numbers at least one is negative.So at least 2018/2=1009 number is negative.
If exactly 1009 is negative then we must have all the number (starting from some) +ve,−ve,+ve,−ve,+ve,−ve,+ve,−ve,.... in this order otherwise we must get two consecutive positive numbers.
So at least 1010 negative number exists.
For construction with 1010 negative take
−x,1,−x+2,1,−x+4,1,−x+6,1,....1,−x+2016,−x+2018 for sufficiently large x.