the answer is $n=9$ $x_{1},x_{2},x_{3},x_{4},x_{5},x_{6},x_{7}=2$,$x_{8}=33,x_{9}=60$ first we know that $ \frac{x-1}{x}\geq \frac{1}{2}$ so we need at least 8 number because $\frac{1}{128}> \frac{15}{2013}$ and now I prove that for n=8 we cant find Such numbers it is obvious that one of our number is at last 61 because if all number smaller than 61 then Denominator is not divisible 61 and now if we have only 8 number the multiplied is at least $(\frac{1}{128})(\frac{60}{61})>\frac{15}{2013}$ so we have at least 9 number

sahadian wrote: the answer is $n=9$ $x_{1},x_{2},x_{3},x_{4},x_{5},x_{6},x_{7}=2$,$x_{8}=33,x_{9}=60$ first we know that $ \frac{x-1}{x}\geq \frac{1}{2}$ so we need at least 8 number because $\frac{1}{128}> \frac{15}{2013}$ and now I prove that for n=8 we cant find Such numbers it is obvious that one of our number is at last 61 because if all number smaller than 61 then Denominator is not divisible 61 and now if we have only 8 number the multiplied is at least $(\frac{1}{128})(\frac{60}{61})>\frac{15}{2013}$ so we have at least 9 number Distinct... Btw, the answer is 134. when $x_1=2$, $x_2=3$, ..., $x_{131}=132$, $x_{132}=181$, $x_{133}=182$, and $x_{134}=183$. Will write down my proof sometime.

WLOG : $x_n>x_{n-1}>...>x_2>x_1\geq 2$ $\Rightarrow x_{i+1}\geq x_i+1$ $\forall i\in\left\{1,2,...,n-1%Error. "Righ" is a bad command. \}$ Using $x_{i+1}\geq x_1+i-1$ and $ x_1\geq2$ we'll get : \[\frac{15}{2013}=\prod_{i=1}^n (1-\frac{1}{x_i})\geq\prod_{i=1}^n ( 1-\frac{1}{x_1+i-1})=\frac{x_1-1}{x_1+n-1}\]\[\Rightarrow 15(x_1+n-1)\geq2013(x_1-1)\Rightarrow n\geq134\] $(x_1,x_2,....,x_{121},x_{122},x_{123},...,x_{134})=(2,3,....,122,131,132,...,143)$ is an example for $n=134$