14. (15 points) From the sequence of consecutive natural numbers $1,2,3, \cdots, 2014$, select $n$ numbers such that no two of these $n$ numbers satisfy the condition where one is 7 times the other. Determine the maximum value of $n$ and provide a justification.
【Analysis】First, group these numbers.
4 numbers in a group: $\{1,7,49,343\},\{2,14,98,686\},\{3,21,147,1029\}\{4,28$, $196,1372\}\{5,35,245,1715\}$
3 numbers in a group: $\{6,42,294\}\{8,56,392\}\{9,63,441\} \cdots\{41,287,2009\}$ Note the first number skips $7, 14, 21, 28, 35$, etc., a total of $41-6+1-5=31$ groups.
2 numbers in a group: $\{43,301\}\{44,308\} \cdots\{286,2002\}$, Note to skip the numbers that have already appeared, i.e., 49, 98, 147, 196, $245, 56, 63, \cdots 280$, a total of 34 numbers, so there are $286-43+1-34=210$ groups of 2 numbers.

【Solution】Solution:
From the analysis, we know: $\{1,7,49,343\},\{2,14,98,686\}, \cdots\{286,2002\}$, in total 246 groups of numbers, in the first five groups, at most 2 can be taken from each group, at least 10 cannot be taken, in the groups of 3 numbers, at least 31 cannot be taken, in the groups of 2 numbers, at least 210 cannot be taken, so the maximum number that can be taken is $2014-10-31-210=1763$ numbers. In these 1763 numbers, answer: taking any 2, one will not be 7 times the other, the maximum value of $n$ is 1763.