No. Taking the sum of all sums of all triples gives $\binom{19}{2}(a_1+a_2+\ldots+a_{20})$ is positive while the sum of all terms is negative. This is a contradiction.

a1+a2+a3>0 a2+a3+a4>0 and so on a18+a19+a20>0 a19+a20+a1>0 a20+a1+a2>0 by adding all these equations, we get a1+a2+.....................+a20>0 which contradicts the question so there does not exist any sequence

This series provides the requirements. -4 -4 9 -4 -4 9 -4 -4 9 -4 -4 9 -4 -4 9 -4 -4 9 -4 -4. The sum of 3 consecutive terms is 1. The sum of all terms is -2 Then answer is YES

No one said it had to be consecutive

Question incorrectly translated. Original says consecutive.

BarisKoyuncu wrote: Question incorrectly translated. Original says consecutive. Thanks. updated.

Isn't it a bit late geo hocam?