Example 7 (2006 National Training Team Selection Contest Question) 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.
Theorem: The largest good number is 576, and the smallest bad number is 443.

Proof:
(1) If positive integers $a, b, c, d$ satisfy $a48^{2}$, then $u+v-2 \geqslant 49$, i.e., $u+v \geqslant 51$, which is 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 $n \leqslant 25$.

Thus, $12(n-1)=13(n+1)-n-25 \geqslant M+50-n-25 \geqslant M$, i.e., $M \leqslant 12(n-1)<13 n \leqslant M+49$.
Therefore, for $1 \leqslant M \leqslant 288$, $M$ is a good number.
Take
\[
\begin{aligned}
\left\{\left(u_{i}, v_{i}\right)\right\}_{i=1}^{23}= & \{(13,26),(14,25),(19,19),(14,26),(15,25),(19,20), \\
& (15,26),(20,20),(17,24),(19,22),(20,21),(13,33), \\
& (18,24),(20,20),(21,21),(15,30),(19,24),(16,29), \\
& (18,26),(19,25),(20,24),(21,23),(14,35)\} .
\end{aligned}
\]

Verification shows that
$u_{i} v_{i} \leqslant\left(u_{i-1}-1\right)\left(v_{i-1}-1\right)+50, i=2,3, \cdots, 23$.
$\left(u_{1}-1\right)\left(v_{1}-1\right)=300, u_{1} v_{1}=338,\left(u_{23}-1\right)\left(v_{23}-1\right)=442$.
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 $\left(u_{i-1}-1\right)\left(v_{i-1}-1\right)<M \leqslant\left(u_{i}-1\right)\left(v_{i}-1\right)$,
$M \leqslant\left(u_{i}-1\right)\left(v_{i}-1\right)<u_{i} v_{i} \leqslant\left(u_{i-1}-1\right)\left(v_{i-1}-1\right)+50 \leqslant M+49, i=2,3, \cdots, 23$.
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
$443 \leqslant(u-1)(v-1)<u v \leqslant 492$.
Thus, $u v-(u-1)(v-1) \leqslant 49$, i.e., $u+v \leqslant 50$.
Also, $443 \leqslant(u-1)(v-1) \leqslant\left(\frac{u+v-2}{2}\right)^{2}$, so $u+v \geqslant 45$.
From $443 \leqslant(u-1)(v-1)=u v-u-v+1 \leqslant u v-2 \sqrt{u v}+1=(\sqrt{u v}-1)^{2}$, we get
$\sqrt{u v} \geqslant 22, u v \geqslant 484$. The values of $u v=484,485,486,487,488,489,490,491,492$ that satisfy $45 \leqslant u+$ $v \leqslant 50$ are only $(u, v)=(14,35),(18,27)$. However, $13 \times 34=442,17 \times 26=442$ contradicts $(u-1)(v-$ $1) \geqslant 443$. Therefore, 443 is the smallest bad number.