For how many natural numbers $n$ between $1$ and $2014$ (both inclusive) is $\frac{8n}{9999-n}$ an integer?

1. We need to determine for how many natural numbers \( n \) between \( 1 \) and \( 2014 \) (both inclusive) the expression \(\frac{8n}{9999-n}\) is an integer.
2. Let \( k = 9999 - n \). Then, the expression becomes \(\frac{8(9999 - k)}{k} = \frac{8 \cdot 9999 - 8k}{k} = \frac{79992 - 8k}{k}\).
3. For \(\frac{79992 - 8k}{k}\) to be an integer, \( 79992 - 8k \) must be divisible by \( k \). This implies that \( 79992 \) must be divisible by \( k \).
4. We need to find the values of \( k \) such that \( 9999 - k \) is between \( 1 \) and \( 2014 \). This translates to \( 7985 \leq k \leq 9998 \).
5. We need to find the divisors of \( 79992 \) that lie within the range \( 7985 \leq k \leq 9998 \).

6. First, we factorize \( 79992 \):
   \[
   79992 = 2^3 \times 3^2 \times 11 \times 151
   \]

7. Next, we check the divisors of \( 79992 \) within the range \( 7985 \leq k \leq 9998 \):
   - Calculate \(\frac{79992}{7985} \approx 10.02\)
   - Calculate \(\frac{79992}{9998} \approx 8.001\)

8. The only integer divisor of \( 79992 \) between \( 8.001 \) and \( 10.02 \) is \( 9 \). Therefore, the only divisor of \( 79992 \) from \( 7985 \) to \( 9998 \) is:
   \[
   \frac{79992}{9} = 8888
   \]

9. This shows that the only possible value of \( k \) is \( 8888 \). Therefore, \( 9999 - n = 8888 \implies n = 9999 - 8888 = 1111 \).

10. Thus, the only natural number \( n \) between \( 1 \) and \( 2014 \) for which \(\frac{8n}{9999-n}\) is an integer is \( 1111 \).

The final answer is \(\boxed{1}\)