Example 7 There is a sequence of numbers: $1,3,4,7,11,18, \cdots$, starting from the third number, each number is the sum of the two preceding numbers.
(1) What is the remainder when the 1991st number is divided by 6?
(2) Group the above sequence as follows:
$(1),(3,4),(7,11,18), \cdots$,

where the $n$-th group has exactly $n$ numbers. What is the remainder when the sum of the numbers in the 1991st group is divided by 6?
(6th Spring Festival Cup Mathematics Competition)
Solution: Let the $n$-th number of the sequence be $a_{n}$. Then
$$
a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3) .
$$

Through experimentation, it is known that the remainders of $a_{1}, a_{2}, \cdots, a_{26}$ when divided by 6 are
$$
\begin{array}{l}
1,3,4,1,5,0,5,5,4,3,1,4,5, \\
3,2,5,1,0,1,1,2,3,5,2,1,3 .
\end{array}
$$

Notice that $a_{25} \equiv a_{1}(\bmod 6), a_{26} \equiv a_{2}(\bmod 6)$, and using $a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3)$, we get
$$
a_{27}=a_{26}+a_{25} \equiv a_{2}+a_{1}=a_{3}(\bmod 6) .
$$

Continuing this way, when $n>24$, we have
$$
a_{n} \equiv a_{n-24}(\bmod 6) \text {. }
$$
(1) Since $1991=24 \times 82+23$, we have
$$
a_{1991} \equiv a_{23} \equiv 5(\bmod 6) \text {. }
$$

Thus, the remainder when the 1991st number is divided by 6 is 5.
(2) When $n>24$, $a_{n} \equiv a_{n-24}(\bmod 6)$, so for any natural number $n$, we have
$$
\begin{array}{l}
a_{n+1}+a_{n+2}+\cdots+a_{n+x} \equiv a_{1}+a_{2}+\cdots+a_{21} \\
\equiv 1+3+4+1+5+0+5+5+4+ \\
3+1+4+5+3+2+5+1+0+ \\
1+1+2+3+5+2 \\
=66 \equiv 0(\bmod 6) .
\end{array}
$$

Since the first 1990 groups contain a total of $1+2+\cdots+1990$ $=1981045$ (let this be $m$) numbers, the sum of the numbers in the 1991st group is
$$
\begin{array}{l}
a_{m+1}+a_{m+2}+\cdots+a_{m+1991} \\
\equiv\left(a_{m+1}+a_{m+2}+\cdots+a_{m+1991}+a_{m+192}\right)- \\
\quad a_{m+192} \\
\equiv 83 \times 0-a_{m+1922} \equiv-a_{13} \equiv-5 \equiv 1(\bmod 6) .
\end{array}
$$

Thus, the remainder when the sum of the numbers in the 1991st group is divided by 6 is 1.