Suppose it is possible to have more than 16 terms in the sequence. Let $a_{1}, a_{2}, \ldots, a_{17}$ be the first 17 terms of the sequence. Consider the following array of terms in the sequence: $$\begin{array}{lllllllllll} a_{1} & a_{2} & a_{3} & a_{4} & a_{5} & a_{6} & a_{7} & a_{8} & a_{9} & a_{10} & a_{11} \ a_{2} & a_{3} & a_{4} & a_{5} & a_{6} & a_{7} & a_{8} & a_{9} & a_{10} & a_{11} & a_{12} \ a_{3} & a_{4} & a_{5} & a_{6} & a_{7} & a_{8} & a_{9} & a_{10} & a_{11} & a_{12} & a_{13} \ a_{4} & a_{5} & a_{6} & a_{7} & a_{8} & a_{9} & a_{10} & a_{11} & a_{12} & a_{13} & a_{14} \ a_{5} & a_{6} & a_{7} & a_{8} & a_{9} & a_{10} & a_{11} & a_{12} & a_{13} & a_{14} & a_{15} \ a_{6} & a_{7} & a_{8} & a_{9} & a_{10} & a_{11} & a_{12} & a_{13} & a_{14} & a_{15} & a_{16} \ a_{7} & a_{8} & a_{9} & a_{10} & a_{11} & a_{12} & a_{13} & a_{14} & a_{15} & a_{16} & a_{17} \end{array}$$ Let $S$ the sum of the numbers in the array. If we sum by rows we obtain negative sums in each row, so $S<0$; however, it we sum by columns we obtain positive sums in each column, so $S>0$, a contradiction. This implies that the sequence cannot have more than 16 terms. One idea to find a suitable sequence with 16 terms is considering cycles of 7 numbers. For instance, one can try $-a,-a, b,-a,-a,-a, b,-a,-a, b,-a,-a,-a, b,-a,-a$. The sum of every seven consecutive numbers is $-5 a+2 b$ and the sum of every eleven consecutive numbers is $-8 a+3 b$, so $-5 a+2 b>0$ and $-8 a+3 b<0$, that is, $\frac{5 a}{2}<b<\frac{8 a}{3} \Longleftrightarrow 15 a<6 b<16 a$. Then we can choose, say, $a=7$ and $105<6 b<112 \Longleftrightarrow b=18$. A valid sequence is then $-7,-7,18,-7,-7,-7,18,-7,-7,18,-7,-7,-7,18,-7,-7$.