jmc

Let there be $T$ teams. For each team, there are $\left(\genfrac{}{}{0ex}{}{n-5}{4}\right)$ different subsets of $9$ players including that full team, so the total number of team-(group of 9) pairs is $$T\left(\genfrac{}{}{0ex}{}{n-5}{4}\right).$$ Thus, the expected value of the number of full teams in a random set of $9$ players is $$\frac{T\left(\genfrac{}{}{0ex}{}{n-5}{4}\right)}{\left(\genfrac{}{}{0ex}{}{n}{9}\right)}.$$ Similarly, the expected value of the number of full teams in a random set of $8$ players is $$\frac{T\left(\genfrac{}{}{0ex}{}{n-5}{3}\right)}{\left(\genfrac{}{}{0ex}{}{n}{8}\right)}.$$ The condition is thus equivalent to the existence of a positive integer $T$ such that $$\frac{T\left(\genfrac{}{}{0ex}{}{n-5}{4}\right)}{\left(\genfrac{}{}{0ex}{}{n}{9}\right)}\frac{T\left(\genfrac{}{}{0ex}{}{n-5}{3}\right)}{\left(\genfrac{}{}{0ex}{}{n}{8}\right)}=1.$$ $$T^2\frac{(n-5)!(n-5)!8!9!(n-8)!(n-9)!}{n!n!(n-8)!(n-9)!3!4!}=1$$ $$T^2=((n)(n-1)(n-2)(n-3)(n-4){)}^2\frac{3!4!}{8!9!}$$ $$T^2=((n)(n-1)(n-2)(n-3)(n-4){)}^2\frac{144}{7!7!8\cdot 8\cdot 9}$$ $$T^2=((n)(n-1)(n-2)(n-3)(n-4){)}^2\frac{1}{4\cdot 7!7!}$$ $$T=\frac{(n)(n-1)(n-2)(n-3)(n-4)}{2^5\cdot 3^2\cdot 5\cdot 7}$$ Note that this is always less than $\left(\genfrac{}{}{0ex}{}{n}{5}\right)$, so as long as $T$ is integral, $n$ is a possibility. Thus, we have that this is equivalent to $$2^5\cdot 3^2\cdot 5\cdot 7|(n)(n-1)(n-2)(n-3)(n-4).$$ It is obvious that $5$ divides the RHS, and that $7$ does iff $n\equiv 0,1,2,3,4\mathrm{mod}7$. Also, $3^2$ divides it iff $n\not\equiv 5,8\mathrm{mod}9$. One can also bash out that $2^5$ divides it in $16$ out of the $32$ possible residues $\mathrm{mod}32$. Using all numbers from $2$ to $2017$, inclusive, it is clear that each possible residue $\mathrm{mod}7,9,32$ is reached an equal number of times, so the total number of working $n$ in that range is $5\cdot 7\cdot 16=560$. However, we must subtract the number of "working" $2\le n\le 8$, which is $3$. Thus, the answer is $\boxed{557}$.