A set of $n$ people participate in an online video basketball tournament. Each person may be a member of any number of $5$-player teams, but no two teams may have exactly the same $5$ members. The site statistics show a curious fact: The average, over all subsets of size $9$ of the set of $n$ participants, of the number of complete teams whose members are among those $9$ people is equal to the reciprocal of the average, over all subsets of size $8$ of the set of $n$ participants, of the number of complete teams whose members are among those $8$ people. How many values $n$, $9\leq n\leq 2017$, can be the number of participants?

$\textbf{(A) } 477 \qquad \textbf{(B) } 482 \qquad \textbf{(C) } 487 \qquad \textbf{(D) } 557 \qquad \textbf{(E) } 562$
1. Let the number of teams be \( T \). We need to determine the averages used in the problem. For the average involving the size \( 9 \) subsets, note that each team is a part of \( \binom{n-5}{4} \) size \( 9 \) subsets, since we need to pick \( 4 \) other participants to be a part of the subset. Thus, this average is 
   \[
   \frac{T \cdot \binom{n-5}{4}}{\binom{n}{9}}.
   \]
2. Similarly, the average for the size \( 8 \) subsets is 
   \[
   \frac{T \cdot \binom{n-5}{3}}{\binom{n}{8}}.
   \]
3. According to the problem, these two averages are reciprocals of each other:
   \[
   \frac{T \cdot \binom{n-5}{4}}{\binom{n}{9}} = \frac{\binom{n}{8}}{T \cdot \binom{n-5}{3}}.
   \]
4. Isolating \( T \) gives:
   \[
   T^2 = \frac{\binom{n}{8} \cdot \binom{n}{9}}{\binom{n-5}{4} \cdot \binom{n-5}{3}}.
   \]
5. Expanding the right-hand side using the explicit formulas for the binomial coefficients and simplifying slightly gives:
   \[
   T^2 = \frac{n^2 (n-1)^2 (n-2)^2 (n-3)^2 (n-4)^2 \cdot 4}{8^2 \cdot 7^2 \cdot 6^2 \cdot 5^2 \cdot 4^2 \cdot 9}.
   \]
6. Simplifying further:
   \[
   T^2 = \left( \frac{2 \cdot n (n-1) (n-2) (n-3) (n-4)}{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3} \right)^2.
   \]
7. Since \( T \) must be an integer, we know that:
   \[
   8 \cdot 7 \cdot 6 \cdot 5 \cdot 2 \cdot 3 \mid n(n-1)(n-2)(n-3)(n-4).
   \]
8. We now determine all possible residues for \( n \) modulo \( 32, 9, 5, 7 \) for which this is an integer.
   - \(\pmod{7}\): One of the 5 terms in the product must be divisible by \( 7 \), so \( n \pmod{7} \in \{0, 1, 2, 3, 4\} \), which is \( 5 \) possibilities.
   - \(\pmod{5}\): One of the 5 terms in the product must be divisible by \( 5 \), but this always occurs, so we do not need to worry about this case.
   - \(\pmod{9}\): \( v_3 \) of the product must be at least \( 2 \), which gives the \( 7 \) possibilities for \( n \pmod{9} \) of \( 0, 1, 2, 3, 4, 6, 7 \).
   - \(\pmod{32}\): \( v_2 \) of the product must be at least \( 5 \), but since there are always at least two even terms in the product, we actually only need to deal with \(\pmod{16}\). There are \( 8 \) possibilities: \( n \pmod{16} \in \{0, 1, 2, 3, 4, 8, 10, 12\} \).

9. By the Chinese Remainder Theorem, we have \( 5 \cdot 7 \cdot 8 = 280 \) possible residues for \( n \) modulo \( 7 \cdot 9 \cdot 16 = 1008 \), giving \( 560 \) possible values for \( n \) in the range \( [1, 2016] \).

10. We then account for the slight difference in the given range:
    - \( 2017 \) is \( 1 \) mod 1008, so it is a valid residue which we must add.
    - Of the numbers from \( 1 \) to \( 8 \), only \( 1, 2, 3, 4, 8 \) work mod 16, and \( 8 \) does not work mod 9, so we must subtract only \( 1, 2, 3, 4 \), all of which clearly work for all three moduli.

11. Thus, we have:
    \[
    560 + 1 - 4 = 557.
    \]

The final answer is \(\boxed{557}\).