Is it integer or not?

We note that for $1 \le n \le 2014$ , $\dfrac {8n}{9999-n}$ is positive and for it to be a positive integer.

$\begin{aligned} \frac {8n}{9999-n} & \ge 1 \\ 8n & \ge 9999 - n \\ 9n & \ge 9999 \\ n & \ge 1111 \end{aligned}$

Therefore, there is no integer solution for $n < 1111$ and when $n = 1111$ , $\dfrac {8n}{9999-n} = 1$ , a solution.

We note that $\dfrac {8n}{9999-n}$ is an increasing function, because:

$\begin{aligned} \frac d{dn} \left(\frac {8n}{9999-n} \right) & = \frac {8(9999-n)+8n}{(9999-n)^2} = \frac {8(9999)}{(9999-n)^2}> 0 \ \forall \ n \end{aligned}$

Since, when $n = 2014$ , $\dfrac {8n}{9999-n} = \dfrac {8(2014)}{9999-2014} \approx 2.018$ , the only other integer solution is 2. Let us check:

$\begin{aligned} \frac {8n}{9999-n} & = 2 \\ 8n & = 19998 - 2n \\ 10n & = 19998 \\ n & = 1999.8 \color{#D61F06} \text{ (Not an integer)} \end{aligned}$

Therefore, there is only $\boxed{1}$ integer solution, that is 1, when $n =$ 1111.