【Analysis and Solution】
(1) $1,3,5,7, \ldots, 2013,2015,2017$; these numbers $\div 2$ have a remainder of 1;
(2) $1,4,7,10, \ldots, 2011,2014,2017$; these numbers $\div 3$ have a remainder of 1;

The numbers that appear in both sequences satisfy both $\div 2$ with a remainder of 1 and $\div 3$ with a remainder of 1, i.e., $\div 6$ with a remainder of 1;
The numbers that appear in both sequences are $1,7,13, 19, \ldots, 2011, 2017$;
The numbers that appear in both sequences total $(2017-1) \div 6+1=337$ numbers.
【Question 6】  
There are the following two sequences of numbers: (1) $1,3,5,7, \cdots, 2013,2015,2017$; (2) 1, 4, 7, 10, , , 2011, 2014, 2017. The numbers that appear in both sequences are $\qquad$ in total.