This problem was proposed by Cyprus

Lukaluce wrote: Find the largest integer $k$ ($k \ge 2$), for which there exists an integer $n$ ($n \ge k$) such that from any collection of $n$ consecutive positive integers one can always choose $k$ numbers, which verify the following conditions: 1. each chosen number is not divisible by $6$, by $7$, nor by $8$; 2. the positive difference of any two distinct chosen numbers is not divisible by at least one of the numbers $6$, $7$, and $8$. Is $max(k)=60?$

The answer is 60. By the second condition given in the question we can understand that, two chosen numbers a and b can't be congruent in modulo 168. That's why we are going to solve this question in modulo 168. In Modulo 168 , because of the first condition, none of the numbers in mod 168 can't be divisible by 6,by 7 nor 8. By doing inclusion -exculision principle we get that there can be at most 60 different numbers in mod 168. For k=60 works when we choose n as 168. 61 doesn't work because from 61 numbers chosen there must be at least two that are congruent in modulo 168. (Considering the first condition)

Claim:- We can't choose more than $168$ numbers. Proof:- Suppose that we can choose more than 168 numbers.Then by the pigeon hole principle,atleast two of them must have a difference of 168 which is a contradiction to the second statement. (To get the maximum number of numbers we choose numbers like 1,2,3,4,5... so if we choose more than 168 numbers then their difference will be 168). Now clearly all numbers except those divisible by $6,7,8$ work hence we use the inclusion-exlusion principle to get the answer to be $60$ For a construction choose $168k+1,168k+2,.....,168k+5,168k+9......$ to be $60$ numbers which satisfy.