Answer: $n=11$. Solution. Obviously, $\mathrm{n} \leq 12$, since there is only one pair of numbers with a difference of 13. Suppose the required is possible for $n=12$. The number 12 can be represented as the difference of numbers from 1 to 14 in exactly two ways: 13-1 and 14-2. Let the number 1 be red for definiteness. Then the number 13 is also red, and the numbers 2 and 14 are blue. Next, there are three pairs with a difference of 11: 12-1, 13-2, 14-3. The pair 13 and 2 is bicolored, so the other two pairs are monochromatic, which means the number 12 is red (like 1), and the number 3 is blue. Continuing in this way to consider differences of $10, 9, 8$, and 7, at each step we will find that all possible pairs, except for two, are already bicolored, and therefore the colors of two more numbers are uniquely determined. In the end, we will find that the numbers from 2 to 7 inclusive are blue, and the numbers from 8 to 13 are red. But in this case, the number 6 cannot be represented as the difference of either red or blue numbers - a contradiction. Therefore, $n \leq 11$. It remains to show that $n$ can equal 11. One example looks like this: the numbers $1,2,4,6,8,10,12$ are red, and the numbers 14, 13, 11, 9, 7, 5, 3 are blue (the arrangement of blue and red numbers is symmetric relative to the midpoint of the segment [1, 14]). There are other examples as well.
7. For what largest $n$ can the numbers $1,2, \ldots, 14$ be colored red and blue so that for any number $k=1,2, \ldots, n$ there exist a pair of blue numbers whose difference is $k$, and a pair of red numbers whose difference is also $k$? (D. Khramtsov)