(14) A finite sequence satisfies: the sum of any 3 consecutive terms is negative, and the sum of any 4 consecutive terms is positive. The maximum number of terms in this sequence is $\qquad$ .
Solution: 5
On the one hand, a sequence of 5 terms can be constructed: $2,2,-5,2,2$ which meets the requirements;
On the other hand, prove that the sequence satisfying the conditions does not exceed 5 terms. Otherwise, take the first 6 terms, and arrange them as follows:
$$
\begin{array}{l}
a_{1}, a_{2}, a_{3} \\
a_{2}, a_{3}, a_{4} \\
a_{3}, a_{4}, a_{5} \\
a_{4}, a_{5}, a_{6}
\end{array}
$$

From the fact that the sum of each row is negative, we know that the sum of these 12 numbers is negative; from the fact that the sum of each column is positive, we know that the sum of these 12 numbers is positive. This is a contradiction.