Solution:

Let $T$ be the number of teams. Summing over size-$9$ subsets counts each team $\binom{n-5}{4}$ times and over size-$8$ subsets $\binom{n-5}{3}$ times. The averages are $\dfrac{\binom{n-5}{4}T}{\binom n9}$ and $\dfrac{\binom{n-5}{3}T}{\binom n8};$ setting the first equal to the reciprocal of the second and simplifying gives $$T = \frac{n(n-1)(n-2)(n-3)(n-4)}{2^5 \cdot 3^2 \cdot 5 \cdot 7}.$$ We need this to be a positive integer with $n \ge 9.$ Let $N = n(n-1)(n-2)(n-3)(n-4);$ as a product of five consecutive integers, $N$ is always divisible by $5.$ Checking residues, $7 \mid N,$ $9 \mid N,$ and $32 \mid N$ each hold for a fixed set of residues, giving $5 \cdot 7 \cdot 8 = 280$ solutions modulo $1008.$ So there are $560$ values in $1 \le n \le 2016;$ removing $n = 1, 2, 3, 4$ (which are below $9$) and adding $n = 2017$ (since $2017 \equiv 1 \pmod{1008}$) gives $560 - 4 + 1 = 557$ valid values.

Thus, the correct answer is D.