For a positive integer $M$, if there exist integers $a, b, c, d$ such that $M \leqslant a<b \leqslant c<d \leqslant M+49, a d=b c$, then $M$ is called a good number; otherwise, $M$ is called a bad number. Try to find the largest good number and the smallest bad number.
Three, the largest good number is 576, and the smallest bad number is 443.
Lemma If positive integers $a, b, c, d$ satisfy $a48^{2}$.
Thus, $u+v-2 \geqslant 49$, i.e., $u+v \geqslant 51$, a contradiction.
Therefore, 576 is the largest good number.
(2) When $1 \leqslant M \leqslant 288$, take an integer $n$ such that
$13 n \leqslant M+49<13(n+1)$.
Then $13 n \leqslant M+49 \leqslant 337$. Thus, $n \leqslant 25$. This way,
$12(n-1)=13(n+1)-n-25$
$\geqslant M+50-n-25 \geqslant M$,

i.e., $M \leqslant 12(n-1)<13 n \leqslant M+49$.
Therefore, when $1 \leqslant M \leqslant 288$, $M$ is a good number.
Take $\left\{\left(u_{i}, v_{i}\right)\right\}_{i=1}^{23}=\{(13,26),(14,25),(19,19)$,
$(14,26),(15,25),(19,20),(15,26),(20,20),(17$,
24), (19, 22), $(20,21),(13,33),(18,24),(20,22)$,
$(21,21),(15,30),(19,24),(16,29),(18,26),(19$,
25), $(20,24),(21,23),(14,35)\}$.
Verification shows that for $i=2,3, \cdots, 23$, we have
$u_{i} v_{i} \leqslant\left(u_{i-1}-1\right)\left(v_{i-1}-1\right)+50$,
$\left(u_{i-1}-1\right)\left(v_{i-1}-1\right)<\left(u_{i}-1\right)\left(v_{i}-1\right)$,
$\left(u_{1}-1\right)\left(v_{1}-1\right)=300, u_{1} v_{1}=338$,
$\left(u_{23}-1\right)\left(v_{23}-1\right)=442$.
When $288<M \leqslant 300$,
$M \leqslant\left(u_{1}-1\right)\left(v_{1}-1\right)<u_{1} v_{1} \leqslant M+49$.
When $\left(u_{i-1}-1\right)\left(v_{i-1}-1\right)<M \leqslant\left(u_{i}-1\right)\left(v_{i}-1\right)$

we have
$M \leqslant\left(u_{i}-1\right)\left(v_{i}-1\right)<u_{i} v_{i}$
$\leqslant\left(u_{i-1}-1\right)\left(v_{i-1}-1\right)+50 \leqslant M+49$,
$i=2,3, \cdots, 23$.
Therefore, when $288 \leqslant M \leqslant 442$, $M$ is a good number.
Next, we prove that 443 is a bad number.
Assume 443 is a good number, then by the lemma, there exist positive integers $u, v$ $(u \leqslant v)$, such that $443 \leqslant(u-1)(v-1)<w \leqslant 492$.
Thus, $w v-(u-1)(v-1) \leqslant 49$, i.e.,
$u+v \leqslant 50$.
Also, $443 \leqslant(u-1)(v-1) \leqslant\left(\frac{u+v-2}{2}\right)^{2}$, so
$u+v \geqslant 45$.
By $443 \leqslant(u-1)(v-1)=u v-u-v+1$
$\leqslant w-2 \sqrt{u v}+1=(\sqrt{u v}-1)^{2}$,

we know $\sqrt{u v} \geqslant 22, u v \geqslant 484$.
Thus, $u v=484,485,486,487,488,489,490,491,492$

satisfying $45 \leqslant u+v \leqslant 50$ are only
$(u, v)=(14,35),(18,27)$.
But $13 \times 34=442,17 \times 26=442$ contradicts $(u-1)(v-1)$ $\geqslant 443$.
Therefore, 443 is the smallest bad number.