There are totally $\binom{2015}{1000}$ subsets. Meanwhile the number of subset whose least element is $k$ equals $\binom{2015-k}{1000-1}$. Thus, the desired result is $$M=\left(1\cdot \binom{2014}{999} + 2\binom{2013}{999} +\cdots + 1016\binom{999}{999}\right) \div \binom{2015}{1000}$$

Setting $n=2014$ and $k=999$ in the result of # 3288 gives the result as $$M=\binom{2016}{1001} \div \binom{2015}{1000} = \frac{2016}{1001}=\boxed{\frac{288}{143}}$$