The set $ \{1,2,3,4, 14,15,16,17, 27,28,29,30, 40,41,42,43, ...\}$ is good (contains exactly $ 372$ numbers) and contains $ 600$. I think this is the only good set, but I don't have the proof. Has anyone?

Yes, your example is in fact the only good set. We count the number of pairs $ (x, I)$, where $ x\in S$ and $ I$ is an interval of 13 consecutive integers containing $ x$. Method 1: For any interval $ I=\{i+1, i+2,\dots i+13\}$, $ S$ can contain at most 1 number from each of $ (i+1,i+5,i+10)$ $ (i+2,i+6,i+11)$ $ (i+3,i+7,i+12)$ $ (i+4,i+8,i+13)$ So at most 4 numbers from $ I$. However, not every interval contain exactly 4 numbers from $ I$. For example, the interval $ (1200, 1201, 1202,\dots 1212)$ can only contain one number from $ S$, and in reality we have $ |I\cap S|\leq\min(|I\cap\{1\dots 1200\}|, 4)$ There are 1212 intervals intersecting $ S$, including 3 "short" intervals on each side whose intersection is less than 4. Adding up over all these intervals gives that there are at most $ 1+2+3+4*1206+1+2+3=4836$ pairs. Method 2: There are 372 integers in $ S$. Each one lies in 13 intervals. Thus there are $ 372*13=4836$ pairs. This implies that equality must hold in our calculations in step 1, and in particular that each interval must contain exactly $ \min(|I\cap\{1,\dots 1200\}|, 4)$ elements of $ S$. This implies that $ \{1,2,3,4\}\subset S$ (by looking at the interval $ \{-8,-7,\dots 4\}$), and that for $ a\leq 1187$ that $ a\in S$ iff $ a+13\in S$ (compare the number of elements of $ S$ in $ \{a,a+1,\dots a+12\}$ to the number in $ \{a+1,\dots a+13\}$).

I will use Dirichlet Principle when I will inshallah prove your identity. Let 600 is not an element of S. b is the first element of S and so b + 4, b + 5, b + 9 are not. The others between them all are not known that whether or not they are elements. So in 4, we have only one good sample. Then we would have only 300 elements in such manner. But we have 372 elements. Then it is guaranttee that there are 72 elements which are compansed each other. So then we can again have 300/4 = 75 and we have 375. 375 - 372 = 3. So let 600 be the one of these 3. But we cannot have. Proven. It is certainly true, if you're claiming that I'm wrong thus you haven't understood the proof.

madshiet wrote: It is certainly true, if you're claiming that I'm wrong thus you haven't understood the proof. Well dang, if Poincare had just said that when he gave his (not correct) proof to his conjecture maybe it wouldn't be a Millennium Problem today. ... There's two things wrong with that statement 1. You're not infallible. Neither is anyone else. 2. A proof isn't any good if no one understands it. As far as the second point is concerned... Quote: But we have 372 elements. Then it is guaranttee that there are 72 elements which are compansed each other. So then we can again have 300/4 = 75 I have absolutely no idea what's going on here.