解析：
$\mid \because a_1=1$, 易得 $a_2=\frac{2}{3} \in(0,1)$, 依次类推可得 $a_n \in(0,1)$
由题意, $a_{n+1}=a_n\left(1-\frac{1}{3} a_n\right)$, 即 $\frac{1}{a_{n+1}}=\frac{3}{a_n\left(3-a_n\right)}=\frac{1}{a_n}+\frac{1}{3-a_n}$,
$$
\therefore \frac{1}{a_{n+1}}-\frac{1}{a_n}=\frac{1}{3-a_n} > \frac{1}{3} \text {, }
$$
即 $\frac{1}{a_2}-\frac{1}{a_1} > \frac{1}{3}, \frac{1}{a_3}-\frac{1}{a_2} > \frac{1}{3}, \frac{1}{a_4}-\frac{1}{a_3} > \frac{1}{3}, \ldots, \frac{1}{a_n}-\frac{1}{a_{n-1}} > \frac{1}{3},(n \geq 2)$,
累加可得 $\frac{1}{a_n}-1 > \frac{1}{3}(n-1)$, 即 $\frac{1}{a_n} > \frac{1}{3}(n+2),(n \geq 2)$,
$\therefore a_n < \frac{3}{n+2},(n \geq 2)$, 即 $a_{100} < \frac{1}{34}, 100 a_{100} < \frac{100}{34} < 3$,
又 $\frac{1}{a_{n+1}}-\frac{1}{a_n}=\frac{1}{3-a_n} < \frac{1}{3-\frac{3}{n+2}}=\frac{1}{3}\left(1+\frac{1}{n+1}\right),(n \geq 2)$,
$$
\therefore \frac{1}{a_2}-\frac{1}{a_1}=\frac{1}{3}\left(1+\frac{1}{2}\right), \frac{1}{a_3}-\frac{1}{a_2} < \frac{1}{3}\left(1+\frac{1}{3}\right), \frac{1}{a_4}-\frac{1}{a_3} < \frac{1}{3}\left(1+\frac{1}{4}\right), \ldots
$$
$$
\frac{1}{a_n}-\frac{1}{a_{n-1}} < \frac{1}{3}\left(1+\frac{1}{n}\right),(n \geq 3),
$$
累加可得 $\frac{1}{a_n}-1 < \frac{1}{3}(n-1)+\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\mathrm{L}+\frac{1}{n}\right),(n \geq 3)$,
$$
\therefore \frac{1}{a_{100}}-1 < 33+\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\mathrm{L}+\frac{1}{99}\right) < 33+\frac{1}{3}\left(\frac{1}{2} \times 4+\frac{1}{6} \times 94\right) < 39 \text {, }
$$
即 $\frac{1}{a_{100}} < 40, \therefore a_{100} > \frac{1}{40}$, 即 $100 a_{100} > \frac{5}{2}$;
综上: $\frac{5}{2} < 100 a_{100} < 3$.