13. (15 points) From the sequence of consecutive natural numbers $1,2,3, \cdots, 2014$, select $n$ numbers such that: any two numbers taken will not have one number being 5 times the other. Find the maximum value of $n$, and explain your reasoning.
【Analysis】First, analyze that the result of 2014 divided by 5 is 402 with a remainder of 4. Therefore, the 5 times of 403 and above numbers are all greater than 2014. Continue reasoning to find the solution.

【Solution】Solution: According to the problem, we have:
$$
2014 \div 5=402 \cdots 4
$$

Then, 5 times 403 is greater than 2014. Therefore, among the numbers from 403 to 2014, there is no number that is 5 times another number. $402 \div 5=80 \cdots 2$ indicates that the number can be 80, and there is no 5 times number.
$80 \div 5=16$, so among the numbers from 17 to 80, there is no 5 times number.
$16 \div 5=3 \cdots 1$. This means that when the value is 1, 2, or 3, there is no 5 times number.
The values can be $1-3, 17-80, 403-2014$
$$
2014-403+1+80-17+1+3-1+1=1679 \text {. }
$$

In summary, the answer is: 1679.