First, one observation: If all four numbers $a,b,c,d$ can be equal, there is no point for solving this problem, so I think there is a mistake in a statement of the problem. I suppose that it's $M\leq a<b\leq c<d\leq M+49$. I found the greatest GOOD number, and I am pretty sure that all numbers not greater then that are GOOD but i can't prove that (for now). Suppose that $b=a+k_1,c=a+k_2$ and $d=a+k_3$ where $0<k_1\leq k_2<k_3<50$. Then from $ad=bc$ we get $k_3-k_2-k_1=\frac{k_1k_2}a$. So $50>k_3>k_1+k_2$, so $max\{k_1+k_2\}=48 \Rightarrow max\{k_1k_2\}=24^2$. The last one implies that $a\leq24^2$, so if $M\geq 24^2+1$ then $a\geq M\geq 24^2+1$ contradiction! If $M=24^2$ there is a solution: $a=24^2,b=c=24^2+24,d=24^2+49$, so greatest GOOD number is $24^2$. For now, this is it, I'll be back soon bye

Well, I guess you're wrong. I assume the condition is $a<b \leq c <d$. If $ad=bc$ we may find $m,n,p,q$ with $a=mn, b=mp, c=nq, d=pq$ then $m<q, n<p$ so $(p-1)(q-1)\geq mn=a$, and the new system $(p-1)(q-1), (p-1)p, (q-1)q, pq$ also satisfies the condition. Thus we have come to the following characterization: A number $M$ is bad if and only if $pq\leq M+49$ implies $(p-1)(q-1)<M$. Now if $M<18^2$ then $M$ id good, because if we let $(k-1)^2<M+49 \leq k^2$ and $r\leq \sqrt{2k-1}$ with $k^2-r^2\leq M+49<k^2-(r-1)^2$ we take $p=k+r, q=k-r$ yielding $(k-1)^2-r^2 \leq M-1 =m+49-50 <k^2-(r-1)^2-50$ so $k+r \geq 25$ which implies $k \geq 19$. Now we have some work to do (I mean $p=15, q=22$ make all numbers from $330$ to $343$ good and so on) till we get the desired number. I think it's 470 or smth close.

I really don't understand what you want to say. I said (and proved) that greatest GOOD number is $24^2=576$ and I found one $a,b,c,d$ which are fulfilling given condition, so obviously is not, like you said,"470, or something close".

The greatest bad number is 443

Alright, let's give this ROBUST problem its due diligence Quote: For a positive integer $M$, if there exists integers $a$, $b$, $c$, $d$ such that $ad=bc$ and \[ M \le a < b \le c < d \le M + 49 \]then we say that $M$ is good, otherwise we say that $M$ is bad. Determine the greatest good number and the smallest bad number. The largest good number is $24^2=576$ and the smallest bad number is $443$. We prove this through a series of several claims. Lemma: The number $M$ is good if and only if: there exist integers $p$ and $q$ for which $(p+1)(q+1) \le M + 49$ but $pq \ge M$. Proof. First, assume $M$ is good. Then given $ad=bc$ we set $a=wx$, $d=yz$, $b=wy$, $c=xz$. Then $a < b$ implies $x < y$, and $b < d$ implies $w < z$. Then $M \le a \le wx \le (z-1)(y-1)$ so we just take $p = z-1$ and $q = y-1$. For the converse direction, if WLOG $p \le q$ then take $(w,x,y,z) = (p,q,q+1,p+1)$ to get $a$, $b$, $c$, $d$. $\blacksquare$ It is now easy to determine the largest good integer. Lemma: The largest good integer is $576 = 24^2$. Proof. To see $576$ is good, take $p=q=24$ in the lemma above. Conversely, assume $M$ is good, so $p$ and $q$ exist as described above. Then we must have $p+q+1 \le 49$ hence $p+q \le 48$. So $M \le pq \le 24^2 = 576$. $\blacksquare$ It remains to determine the smallest bad integer. First, we contend that: Lemma: Every integer $M \le 288$ is good. Proof. Note that there is some multiple of $13$ in $\{M+37, M+38, \dots, M+49\}$, call it $K$. We claim that we can take $q = 12$, and $p = K/13-1$. Indeed, \[ pq = \frac{12}{13}K - 12 \ge \frac{12}{13} (M+37) - 12 = M + \frac{12 \cdot 24 - M}{13} \ge M \]as desired. $\blacksquare$ Lemma: Every integer $287 \le M \le 442$ is good. Proof. Note that any pair $(p,q)$ of integers is a witness to all $pq - \delta \le M \le pq$ being prime, where $\delta = 48-p-q$. In light of this, we construct the following 24 cases: \[ \begin{array}{cccc} p \cdot q & pq & \delta & pq-\delta \\ \hline 15 \cdot 20 & 300 & 13 & 287 \\ 14 \cdot 22 & 308 & 12 & 296 \\ 15 \cdot 21 & 315 & 12 & 303 \\ 18 \cdot 18 & 324 & 12 & 312 \\ \hline 15 \cdot 22 & 330 & 11 & 319 \\ 18 \cdot 19 & 342 & 11 & 331 \\ \hline 14 \cdot 25 & 350 & 9 & 341 \\ 19 \cdot 19 & 361 & 10 & 351 \\ \hline 14 \cdot 26 & 364 & 8 & 356 \\ 17 \cdot 22 & 374 & 9 & 365 \\ 19 \cdot 20 & 380 & 9 & 371 \\ \hline 16 \cdot 24 & 384 & 8 & 376 \\ 13 \cdot 30 & 390 & 5 & 385 \\ 18 \cdot 22 & 396 & 8 & 388 \\ 20 \cdot 20 & 400 & 8 & 392 \\ \hline 17 \cdot 24 & 408 & 7 & 401 \\ 18 \cdot 23 & 414 & 7 & 407 \\ 16 \cdot 26 & 416 & 6 & 410 \\ 20 \cdot 21 & 420 & 7 & 413 \\ \hline 17 \cdot 25 & 425 & 6 & 419 \\ 18 \cdot 24 & 432 & 6 & 426 \\ 15 \cdot 29 & 435 & 4 & 431 \\ 21 \cdot 21 & 441 & 6 & 435 \\ \hline 17 \cdot 26 & 442 & 5 & 437 \end{array} \]For readability we have placed dividing lines under any pair with $|p-q| \le 1$. Since the intervals $[pq-\delta, pq]$ cover $[287, 442]$, the lemma is proved. $\blacksquare$ Lemma: The number $M = 443$ is bad. Proof. Assume for contradiction $pq$ exists, meaning $pq \ge 443$ and $(p+1)(q+1) \le 492$. Then $pq \le 491 - (p+q)$. Now $p+q \ge 2\sqrt{443} \implies p+q \ge 43$, hence $pq \le 448$. Hence $K = pq \in [443, 448]$. We now compute the factorization of each $K$ with $p+q$ minimal (in other words $p$ and $q$ as close as possible). \begin{align*} 443 &= 1 \cdot 442 \\ 444 &= 12 \cdot 37 \\ 445 &= 5 \cdot 89 \\ 446 &= 2 \cdot 233 \\ 447 &= 3 \cdot 149 \\ 448 &= 16 \cdot 28 \end{align*}All of these fail the inequality $(p+1)(q+1) \le 492$, so $443$ is bad. $\blacksquare$

The greatest good number is $576$ and the smallest bad number is $443$. Claim: $M$ is good iff there exist positive integers $x \le y$ (WLOG) such that $xy \ge M$ and $(x+1) (y+1) \le M + 49$ (or in other words $(x+1) (y+1) - 49 \le M \le xy$). Proof: The if direction is true by considering $a = xy, b = (x+1) y, c = x(y+1)$, and $d = (x+1)(y+1)$. Now we prove the only if direction. Suppose such $x,y$ do not exist, but we had $a, b, c, d$ where $ad = bc$ and $M \le a < b \le c < d \le M + 49$. By Four Number Lemma, there exist positive integers $p,q,r,s$ with $pq = a, rs = d, pr = b, qs = c$. We must have $q < r$ and $p < s$. Therefore $rs \ge (p+1) (q+1)$, so $pq \ge M$ and $(p+1) (q+1) \le M + 49$, as desired. $\square$ Clearly $576$ is good by setting $x = y = 24$. If $M$ is good, then $xy \ge M$ and $(x+1) (y+1) \le M + 49$ for some $x,y$, so $x + y \le 48$, meaning the maximal value for $xy$ is $24\cdot 24 = 576$, hence all good numbers are at most $576$, so $576$ is the largest good number. Now it remains to prove that $443$ is the smallest bad number. Claim: All $M \le 290$ are good. Proof: Let $10y$ denote the smallest multiple of $10$ at least $M$. Setting $x = 10$ and $y = y$ gives that if $11(y+1) - 49 \le M \le 10y$ implies $M$ is good. Let $M = 10y - c $, where $0\le c \le 9$. We have $11 y - 38 \le 10y - c$, so $y \le (38 - c)$. Since $c$ is at most $9$, any $y \le 29$ works, meaning any $M \le 290$ is good. $\square$ Now we prove that every integer in $[291,442]$ is good. For $[291,300]$ take $(x,y) = (15,20)$. For $[301,311]$ take $(x,y) = (13,24)$. For $[312,324]$ take $(x,y) = (18,18)$. For $[325,330]$ take $(x,y) = (15,22)$. For $[331,340]$ take $(x,y) = (17,20)$. For $[341,350]$ take $(x,y) = (14,25)$. For $[351,361]$ take $(x,y) = (19,19)$. For $[362,368]$ take $(x,y) = (16,23)$. For $[369,378]$ take $(x,y) = (18,21)$. For $[379,384]$ take $(x,y) = (16,24)$. For $[385,390]$ take $(x,y) = (15,26)$. For $391$ take $(x,y) = (15,28)$. For $[392,400]$ take $(x,y) = (20,20)$. For $[401,408]$ take $(x,y) = (17,24)$. For $[409,414]$ take $(x,y) = (18,23)$. For $[415,420]$ take $(x,y) = (20,21)$. For $[421,425]$ take $(x,y) = (17,25)$. For $[426,432]$ take $(x,y) = (18,24)$. For $[433,434]$ take $(x,y) = (14,31)$. For $[435,441]$ take $(x,y) = (21,21)$. For $442$ take $(x,y) = (17,26)$. Therefore every positive integer less than $443$ is good. It remains to show that $443$ is bad. Suppose there existed $x,y$ where $(x+1)(y+1) - 49 \le 443 \le xy$. Let $xy = 443 + c$. We must have $xy + x + y - 48 \le xy - c \le xy$, so $x + y \le 48 - c \le 48$. However, since $xy \ge 443$, we have $x + y \ge 2\sqrt{xy} > 42$, so $c < 6$. Now it remains to disprove $c\in [0,5]$. If $c=0$, then $xy = 443$. To check that $443$ is prime, we see it is $420 + 23$, so it can't be divisible by $2,3,5,7$. It is also $143 + 300$, so it isn't divisible by $11$ or $13$, and it is $323 + 120$, so it isn't divisible by $17$ or $19$. If $443$ was composite, then it would have a prime factor under $\sqrt{443} < 22$, but we proved this is impossible. Thus, we must have $(x,y) = (1,443)$ and permutations, so $x + y = 444 > 48$, absurd. If $c = 1$, then $xy = 444$ and $x + y \le 48$. However, one of $x,y$ must be a multiple of $37$ since $37 \mid 444$ and $37$ is prime. If the multiple of $37$ is over $74$, then it's obviously impossible. Thus, one of $x,y$ is equal to $37$, so the other one must be $12$, meaning $x + y = 49 > 48$, absurd. If $c = 2$, then $xy = 445$ and $x + y \le 48 - c = 46$. Then since $\frac{445}{5} = 89$ is prime, one of $x,y$ is a multiple of $89$, so $x + y > 89$, absurd. If $c = 3$, then $xy = 446$, and $x + y \le 45$. We see that $\frac{446}{2} = 223$ is prime because it is $2\cdot 3\cdot 5\cdot 7 + 13$, $11\cdot 20 + 3$, $13\cdot 17 + 2$ so it can't be a multiple of any prime under $\sqrt{223}$. Therefore, one of $x,y$ is a multiple of $223$, so $x + y \ge 223$, absurd. If $c = 4$, then $xy = 447$ and $x + y \le 44$. Since $\frac{447}{3} = 149$ is prime, one of $x,y$ is a multiple of $149$, so $x + y \ge 149$, absurd. If $ c= 5$, then $xy = 448$ and $x + y \le 43$. Clearly $7\mid 448$, so one of $(x,y)$ is a multiple of $7$. If one of $(x,y)$ is $7$ or $14$, then either $x + y = 7 + 64$ or $14 + 32$, both of which exceed $43$. Since $21\nmid 448$, neither of $x,y$ can equal $21$. Now if one of $x,y$ is $28$, then $x + y = 28 + 16 = 44 > 43$. Otherwise, one of $x,y$ is a multiple of $7$ over $42$ (since $35$ and $42$ don't divide $448$), so $x + y > 43$, absurd.