A student at Harvard named Kevin
Was counting his stones by $11$
He messed up $n$ times
And instead counted $9$s
And wound up at $2007$.

How many values of $n$ could make this limerick true?
1. Let \( m \) be the number of times Kevin correctly counts by \( 11 \), and let \( n \) be the number of times he incorrectly counts by \( 9 \). The total count he ends up with is given by the equation:
   \[
   11m + 9n = 2007
   \]

2. To find the possible values of \( n \), we first reduce the equation modulo \( 11 \):
   \[
   11m + 9n \equiv 2007 \pmod{11}
   \]
   Since \( 11m \equiv 0 \pmod{11} \), we have:
   \[
   9n \equiv 2007 \pmod{11}
   \]

3. Next, we need to find \( 2007 \mod 11 \). We do this by performing the division:
   \[
   2007 \div 11 = 182 \quad \text{with a remainder of} \quad 2007 - 182 \times 11 = 2007 - 2002 = 5
   \]
   Therefore:
   \[
   2007 \equiv 5 \pmod{11}
   \]

4. Substituting this back into our congruence, we get:
   \[
   9n \equiv 5 \pmod{11}
   \]

5. To solve for \( n \), we need the multiplicative inverse of \( 9 \) modulo \( 11 \). We find this by solving:
   \[
   9x \equiv 1 \pmod{11}
   \]
   By trial, we find that:
   \[
   9 \times 5 = 45 \equiv 1 \pmod{11}
   \]
   Thus, the inverse of \( 9 \) modulo \( 11 \) is \( 5 \).

6. Multiplying both sides of the congruence \( 9n \equiv 5 \pmod{11} \) by \( 5 \), we get:
   \[
   n \equiv 5 \times 5 \pmod{11} \implies n \equiv 25 \pmod{11} \implies n \equiv 3 \pmod{11}
   \]

7. Therefore, \( n \) can be written as:
   \[
   n = 11k + 3 \quad \text{for integer} \quad k
   \]

8. To find the range of possible values for \( n \), we note that \( n \) must be a non-negative integer such that \( 11m + 9n = 2007 \) is satisfied. Since \( 9n \leq 2007 \), we have:
   \[
   n \leq \frac{2007}{9} \approx 223
   \]

9. We need to find the number of integers \( k \) such that:
   \[
   0 \leq 11k + 3 \leq 223
   \]
   Solving for \( k \):
   \[
   0 \leq 11k + 3 \leq 223 \implies -3 \leq 11k \leq 220 \implies 0 \leq k \leq 20
   \]

10. Therefore, \( k \) can take on \( 21 \) values (from \( 0 \) to \( 20 \)).

The final answer is \(\boxed{21}\).