$2 \cdot 10$ In the known sequence $1,4,8,10,16,19,21,25,30,43$, how many groups of consecutive numbers have a sum that is divisible by 11.
(China High School Mathematics League, 1985)
[Solution] Let the corresponding terms of the sequence be $a_{i}, i=1,2, \cdots, 10$. And let
$$
S_{k}=a_{1}+a_{2}+\cdots+a_{k} \text {. }
$$

From this, the sequence $S_{1}, S_{2}, \cdots, S_{10}$ can be calculated as
$$
1,5,13,23,39,58,79,104,134,177 \text {. }
$$

Their remainders when divided by 11 are
$$
1,5,2,1,6,3,2,5,2,1 \text {. }
$$

From this, we can obtain
$$
\begin{array}{ll}
S_{1} \equiv S_{4} & (\bmod 11) \\
S_{1} \equiv S_{10} & (\bmod 11) \\
S_{4} \equiv S_{10} & (\bmod 11) \\
S_{2} \equiv S_{8} & (\bmod 11) \\
S_{3} \equiv S_{7} & (\bmod 11) \\
S_{7} \equiv S_{9} & (\bmod 11) \\
S_{3} \equiv S_{9} & (\bmod 11)
\end{array}
$$

Since $S_{k}-S_{j}(k>j)$ is the sum of consecutive terms of the sequence $\left\{a_{i}\right\}$, and when $S_{k} \equiv$ $S_{j}(\bmod 11)$, $11 \mid S_{k}-S_{j}$, there are a total of 7 groups of arrays that meet the requirements of the problem.