3.  There are 10 positive integers arranged from smallest to largest: $1, 4, 8$, $10, 16, 19, 21, 25, 30, 43$. How many groups of consecutive numbers have a sum that is divisible by 11?
Let 10 positive integers be $a_{1}, a_{2}$, $\cdots, a_{10}$, and let $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$. Then
$$
a_{i+1}+a_{i+2}+\cdots+a_{j}=S_{j}-S_{i} .
$$

The sequence $a_{n}(n=1,2, \cdots, 10)$ has remainders modulo 11 of $1,4,-3,-1,5,-3,-1,3,-3,-1$; the sum of the first $n$ terms $S_{n}$ has remainders modulo 11 of $1,5,2$, $1,6,3,2,5,2,1$. Therefore,
$$
\begin{array}{l}
S_{1} \equiv S_{4} \equiv S_{10} \equiv 1(\bmod 11), \\
S_{2} \equiv S_{8} \equiv 5(\bmod 11), \\
S_{3} \equiv S_{7} \equiv S_{9} \equiv 2(\bmod 11) .
\end{array}
$$

Thus, there are $3+1+3=7$ pairs $(i, j)$ such that $11 \mid \left(S_{j}-S_{i}\right)$, i.e., $11 \mid \left(a_{i+1}+a_{i+2}+\cdots+a_{j}\right)$.