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 $x$.