Find the largest integer $k(k \geq 2)$, for which there exists an integer $n(n \geq k)$ 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 and by 8;
2. the positive difference of any two different chosen numbers is not divisible by at least one of the numbers 6, 7 or 8.
An integer is divisible by 6, 7, and 8 if and only if it is divisible by their Least Common Multiple, which equals $6 \times 7 \times 4 = 168$.

Let $n$ be a positive integer and let $A$ be an arbitrary set of $n$ consecutive positive integers. Replace each number $a_{i}$ from $A$ with its remainder $r_{i}$ (mod 168). The number $a_{i}$ is divisible by 6 (7 or 8) if and only if its remainder $r_{i}$ is divisible by 6 (respectively 7 or 8). The difference $\left|a_{i} - a_{j}\right|$ is divisible by 168 if and only if their remainders $r_{i} = r_{j}$.

Choosing $k$ numbers from the initial set $A$, which verify the required conditions, is the same as choosing $k$ their remainders (mod 168) such that:

1. each chosen remainder is not divisible by 6, 7, and 8;
2. all chosen remainders are different.

Suppose we have chosen $k$ numbers from $A$, which verify the conditions. Therefore, all remainders are different and $k \leq 168$ (otherwise, there would be two equal remainders).

Denote by $B = \{0, 1, 2, 3, \ldots, 167\}$ the set of all possible remainders (mod 168) and by $B_{m}$ the subset of all elements of $B$, which are divisible by $m$. Compute the number of elements of the following subsets:

$$
\begin{gathered}
\left|B_{6}\right| = 168 \div 6 = 28, \quad \left|B_{7}\right| = 168 \div 7 = 24, \quad \left|B_{8}\right| = 168 \div 8 = 21 \\
\left|B_{6} \cap B_{7}\right| = \left|B_{42}\right| = 168 \div 42 = 4, \quad \left|B_{6} \cap B_{8}\right| = \left|B_{24}\right| = 168 \div 24 = 7 \\
\left|B_{7} \cap B_{8}\right| = \left|B_{56}\right| = 168 \div 56 = 3, \quad \left|B_{6} \cap B_{7} \cap B_{8}\right| = \left|B_{168}\right| = 1
\end{gathered}
$$

Denote by $D = B_{6} \cup B_{7} \cup B_{8}$, the subset of all elements of $B$, which are divisible by at least one of the numbers 6, 7, or 8. By the Inclusion-Exclusion principle we got

$$
\begin{gathered}
|D| = \left|B_{6}\right| + \left|B_{7}\right| + \left|B_{8}\right| - \left(\left|B_{6} \cap B_{7}\right| + \left|B_{6} \cap B_{8}\right| + \left|B_{7} \cap B_{8}\right|\right) + \left|B_{6} \cap B_{7} \cap B_{8}\right| = \\
28 + 24 + 21 - (4 + 7 + 3) + 1 = 60.
\end{gathered}
$$

Each chosen remainder belongs to the subset $B \setminus D$, since it is not divisible by 6, 7, and 8. Hence, $k \leq |B \setminus D| = 168 - 60 = 108$.

Let us show that the greatest possible value is $k = 108$. Consider $n = 168$. Given any collection $A$ of 168 consecutive positive integers, replace each number with its remainder (mod 168). Choose from these remainders 108 numbers, which constitute the set $B \setminus D$. Finally, take 108 numbers from the initial set $A$, having exactly these remainders. These $k = 108$ numbers verify the required conditions.