Suppose that this sequence has 17 terms. Then we will have that $a_1+a_2+a_3+\cdot\cdot\cdot +a_{11}<0$ $a_2+a_3+a_4+\cdot\cdot\cdot +a_{12}<0$ $a_3+a_4++a_5\cdot\cdot\cdot +a_{13}<0$ $\cdot\cdot\cdot$ $a_7+a_8+a_9+\cdot\cdot\cdot +a_{17}<0$ And if we add this seven inequalities we will have a sum that is negative, but each column is positive, so the contradiction shows that the terms of this sequence are at most 16. And here is an example $-5, -5, 13, -5, -5, -5, 13, -5, -5, 13, -5, -5, -5, 13, -5, -5$

It's weird that this problem is almost the same as IMO 1977: http://www.mathlinks.ro/viewtopic.php?t=61033

Beat wrote: Suppose that this sequence has 17 terms. Then we will have that $ a_1 + a_2 + a_3 + \cdot\cdot\cdot + a_{11} < 0$ $ a_2 + a_3 + a_4 + \cdot\cdot\cdot + a_{12} < 0$ $ a_3 + a_4 + + a_5\cdot\cdot\cdot + a_{13} < 0$ $ \cdot\cdot\cdot$ $ a_7 + a_8 + a_9 + \cdot\cdot\cdot + a_{17} < 0$ And if we add this seven inequalities we will have a sum that is negative, but each column is positive, so the contradiction shows that the terms of this sequence are at most 16. And here is an example $ - 5, - 5, 13, - 5, - 5, - 5, 13, - 5, - 5, 13, - 5, - 5, - 5, 13, - 5, - 5$ Quite simple and nice ! Congrats