$5 \cdot 101$ Give 2000 people sequential numbers as $1,2, \cdots, 2000$, if the sum of the numbers of some people (can be 1 person, but not none) is a multiple of 5, then it is called a “good group”. Find the number of all possible “good groups” that can be formed from these 2000 people.
【Sol】Let
$$
S_{n}=\{1,2, \cdots, n\} .
$$

For the subsets of $S_{n}$, classify them into 5 categories based on the sum of their elements modulo 5: $M_{j}, j=0,1,2,3$, and let the number of subsets in these 5 categories be $a_{n}, b_{n}, c_{n}, d_{n}, e_{n}$, respectively. Clearly, the problem is to find $a_{2000}$.
By definition, we have
$$
a_{n}+b_{n}+c_{n}+d_{n}+e_{n}=2^{n}, n=0,1,2, \cdots,
$$

and
$$
\begin{array}{l}
a_{5 n}=2 a_{5 n-1}, a_{5 n-1}=a_{5 n-2}+b_{5 n-2}, \\
a_{5 n-2}=a_{5 n-3}+c_{5 n-3}, b_{5 n-2}=b_{5 n-3}+d_{5 n-3} .
\end{array}
$$

Thus, we have
$$
\begin{aligned}
a_{5 n} & =2\left(a_{5 n-3}+b_{5 n-3}+c_{5 n-3}+d_{5 n-3}\right) \\
& =2\left(2^{5 n-3}-e_{5 n-3}\right) .
\end{aligned}
$$

Also,
$$
\begin{array}{l}
e_{5 n-3}=e_{5 n-4}+c_{5 n-4}, e_{5 n-4}=e_{5 n-5}+d_{5 n-5}, \\
c_{5 n-4}=c_{5 n-5}+b_{5 n-5} .
\end{array}
$$

Substituting into (2) yields
$$
\begin{aligned}
a_{5 n} & =2\left(2^{5 n-3}-b_{5 n-5}-c_{5 n-5}-d_{5 n-5}-e_{5 n-5}\right) \\
& =2\left(2^{5 n-3}-2^{5 n-5}+a_{5 n-5}\right) \\
& =3 \times 2^{5 n-4}+2 a_{5 n-5} .
\end{aligned}
$$

Using (3) recursively, we get
$$
\begin{aligned}
a_{5 n} & =3 \times 2^{5 n-4}+3 \times 2^{5 n-8}+4 a_{5 n-10} \\
& =3 \times\left(2^{5 n-4}+2^{5 n-8}+\cdots+2^{5 n-4 n}\right)+2^{n} a_{0} \\
& =3 \times 2^{n} \frac{2^{4 n}-1}{2^{4}-1}+2^{n} .
\end{aligned}
$$

In particular, setting $n=400$ in (4), we obtain
$$
a_{2000}=2^{400}\left[\frac{1}{5}\left(2^{1600}-1\right)+1\right] \text {. }
$$

According to the problem, the empty set is not counted, so the total number of excellent groups is $a_{2000}-1=$ $2^{400}\left[\frac{1}{5}\left(2^{1600}-1\right)+1\right]-1$.