24.1.3 ** In a finite sequence of real numbers, the sum of any 3 consecutive terms is negative, and the sum of any 4 consecutive terms is positive. Find the maximum number of terms \( r \) in this sequence.
Parse, the maximum value of $r$ is 5.
On one hand, construct a 5-term sequence: $2,2,-5,2,2$, where the sum of any 3 consecutive terms is -1, and the sum of any 4 consecutive terms is 1. Therefore, when $r=5$, there exists a 5-term sequence that satisfies the conditions. On the other hand, prove that for any finite sequence (of $r$ terms) that satisfies the conditions, then $r \leqslant 5$. Use proof by contradiction. Assume $r \geqslant 6$, we take the first 6 terms of this sequence, and arrange them in a $4 \times 3$ table as follows:
$$
\begin{array}{l}
a_{1}, a_{2}, a_{3}, \\
a_{2}, a_{3}, a_{4}, \\
a_{3}, a_{4}, a_{5}, \\
a_{4}, a_{5}, a_{6} .
\end{array}
$$

In the above table, each row is the sum of 3 consecutive terms, and each column is the sum of 4 consecutive terms. By the given conditions, the sum of each row is less than 0, and the sum of each column is greater than 0.
Consider calculating the sum of all numbers in the table $M$ (i.e., $a_{1}+2 a_{2}+3 a_{3}+3 a_{4}+2 a_{5}+a_{6}$). Considering each row, the sum of each row is less than 0, so $M<0$. (2)
(1) and (2) are contradictory! Therefore, the assumption is false, hence $r \leqslant 5$.