Maximum value of n?

The problem is to find the largest $n$ , which is the same as finding the smallest $a_i$ for each $i^{th}$ term, so that we have the maximum number of terms, that is the largest $n$ .

Starting from $a_1$ , its smallest value is $1$ , then that of $a_2$ is $3$ , $a_3$ is $5$ and so on. So the longest possible (hence the largest $n$ ) series is:

$\frac {1}{1} < \frac {3}{2} < \frac {5}{3} < ... \frac {2i-1}{i} ... < \frac {2015}{n}$

The largest $n = \dfrac{2015+1}{2} = \boxed{1008}$

To elaborate on Chew's solution: we need to prove that if the ith term is $\frac{2i-1}{i}$ then the next term is at least $\frac{2i+1}{i+1}$ . Now, $\frac{(2i-1)(i+1)}{i} = \frac{2i^2+i-1}{i}=2i+1-\frac{1}{i}$ . It is clear that 2i+1 is the smallest integer larger than this.