$7 \cdot 114$ In a finite sequence of real numbers, the sum of any 7 consecutive terms is negative, while the sum of any 11 consecutive terms is positive. How many terms can such a sequence have at most?
[Sol] If the sequence has 17 terms, it can be arranged as follows:
$$
\begin{array}{c}
a_{1}, a_{2}, a_{3}, \cdots, a_{10}, a_{11}, \\
a_{2}, a_{3}, a_{4}, \cdots, a_{11}, a_{12}, \\
a_{3}, a_{4}, a_{5}, \cdots, a_{12}, a_{13}, \\
\vdots \quad \vdots \quad \vdots \quad \vdots \quad \vdots \\
a_{7}, a_{8}, a_{9}, \cdots, a_{16}, a_{17} .
\end{array}
$$

By the given condition, the sum of the numbers in each row is positive, so the sum of all the numbers in the table is positive; on the other hand, the sum of the numbers in each column is negative, so the sum of all the numbers in the table is negative, which is a contradiction. This indicates that the sequence satisfying the requirement has at most 16 terms.
Consider the following sequence:
$$
5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5 \text {. }
$$

In any 7 consecutive terms, there are 5 fives and two -13s, with a sum of -1; in any 11 consecutive terms, there are 8 fives and 3 -13s, with a sum of 1. It is evident that this sequence meets the requirements of the problem and has 16 terms, hence the sequence satisfying the requirements of the problem has at most 16 terms.