For a positive integer $M$, if there exist integers $a, b, c, d$ such that $M \leqslant a < b \leqslant c < d \leqslant M+49, ad = bc$, then $M$ is called a good number; otherwise, $M$ is called a bad number. Find the largest good number and the smallest bad number. (2006 National Team Selection Examination Question)
The largest good number is 576, and the smallest bad number is 443.
$M$ is a good number if and only if there exist positive integers $u, v$ such that $uv \geqslant M, (u+1) \cdot (v+1) \leqslant M+49(*)$.

Sufficiency: Suppose $u, v$ exist, without loss of generality, assume $u \leqslant v$, then $M \leqslant uv < u(v+1) \leqslant v(u+1) < (u+1)(v+1) \leqslant M+49$, taking $a=uv, b=u(v+1), c=v(u+1), d=(u+1)(v+1)$ is sufficient $(ad=bc$ is obvious).

Necessity: Suppose $a, b, c, d$ exist, from $ad=bc$ we know $\frac{a}{b}=\frac{c}{d}$, let its simplest fraction form be $\frac{u}{r}$, then we can set $a=uv, b=rv, c=us, d=rs$. From $a<b$ we know $u<r, r \geqslant u+1$, from $a<c$ we know $v<s, s \geqslant v+1$.

Therefore, $uv=a \geqslant M, (u+1)(v+1) \leqslant rs=d \leqslant M+49$.
Now, find the largest good number. If $M$ is a good number, then from $(*)$ and the Cauchy-Schwarz inequality, we know
$$\sqrt{M+49} \geqslant \sqrt{(u+1)(v+1)} \geqslant \sqrt{uv}+1 \geqslant \sqrt{M}+1,$$

Thus, $M \leqslant 576$. When $M=576$, taking $u=v=24$ is sufficient, so the largest good number is 576.
Next, find the smallest bad number. First, we can prove that 443 is a bad number.
Actually, assuming for $M=443$ there exist $u, v$ such that $(*)$ holds, then
$$\sqrt{492} \geqslant \sqrt{(u+1)(v+1)} \geqslant \sqrt{uv}+1$$

Thus, $uv \leqslant (\sqrt{492}-1)^2 = 493 - 2\sqrt{492} < 493 - 2\sqrt{484} = 449$, so $443 \leqslant uv \leqslant 448$.

Moreover, when $uv$ is fixed, it is well-known that $u, v$ are closer, $u+v$ is smaller, $(u+1)(v+1) = uv + u + v + 1$ is also smaller.

Set $uv=443=1 \cdot 443$, then $u+v \geqslant 444, (u+1)(v+1) \geqslant 888$;
Set $uv=444=12 \cdot 37$, then $u+v \geqslant 49, (u+1)(v+1) \geqslant 494$;
Set $uv=445=5 \cdot 89$, then $u+v \geqslant 94, (u+1)(v+1) \geqslant 540$;
Set $uv=446=2 \cdot 223$, then $u+v \geqslant 225, (u+1)(v+1) \geqslant 672$;
Set $uv=447=3 \cdot 149$, then $u+v \geqslant 152, (u+1)(v+1) \geqslant 600$;
Set $uv=448=16 \cdot 28$, then $u+v \geqslant 44, (u+1)(v+1) \geqslant 493$.
All of these contradict $(u+1)(v+1) \leqslant 492$, so 443 is indeed a bad number.
Next, prove that all positive integers less than 443 are good numbers.
When $M$ belongs to the following intervals: [245, 258]; [259, 265]; [266, 274]; [275, 280]; [281, 292]; [293, 300]; [301, 311]; [312, 322]; [323, 328]; [329, 334]; [335, 341]; [342, 350]; [351, 358]; [359, 366]; [367, 375]; [376, 382]; [383, 385]; [386, 391]; [392, 400]; [401, 406]; [407, 412]; [413, 418]; [419, 425]; [426, 430]; [413, 433]; [434, 436]; [437, 442], the corresponding $(u, v)$ are $(13,20); (13,21); (14,20); (17,17); (14,21); (17,18); (13,24); (18,18); (11,30); (17,20); (17,24); (18,23); (20,21); (17,25); (18,24); (14,31); (20,22); (17,26)$.

Finally, for $1 \leqslant M \leqslant 245$, set $t^2 \leqslant M < (t+1)^2$, then $1 \leqslant t \leqslant 15$. If $t^2 \leqslant M < t(t+1)$, then take $u=t, v=t+1$, we have $uv \geqslant M, (u+1)(v+1) - M \leqslant (t+1)(t+2) - t^2 = 3t + 2 \leqslant 47$; if $t(t+1) \leqslant M < (t+1)^2$, then take $u=v=t+1$, we have $uv \geqslant M, (u+1)(v+1) - M \leqslant (t+2)^2 - t(t+1) = 3t + 4 \leqslant 49$. In summary, the largest good number is 576, and the smallest bad number is 443.