Let $m$ be the maximum possible value of $x^{16} + \frac{1}{x^{16}}$, where \[x^6 - 4x^4 - 6x^3 - 4x^2 + 1=0.\] Find the remainder when $m$ is divided by $2007$.
To find the maximum possible value of \( x^{16} + \frac{1}{x^{16}} \) given the equation \( x^6 - 4x^4 - 6x^3 - 4x^2 + 1 = 0 \), we will analyze the roots of the polynomial and their properties.

1. **Factor the polynomial:**
   \[
   x^6 - 4x^4 - 6x^3 - 4x^2 + 1 = (x+1)^2 (x^2 - 3x + 1) (x^2 + x + 1)
   \]
   This factorization helps us identify the possible values of \( x \).

2. **Analyze the roots:**
   - For \( x = -1 \):
     \[
     x^{16} + \frac{1}{x^{16}} = (-1)^{16} + \frac{1}{(-1)^{16}} = 1 + 1 = 2
     \]
   - For \( x^2 + x + 1 = 0 \):
     The roots are the non-real cube roots of unity, \( \omega \) and \( \omega^2 \), where \( \omega^3 = 1 \) and \( \omega \neq 1 \). For these roots:
     \[
     x^3 = 1 \implies x^{16} = x \implies x^{16} + \frac{1}{x^{16}} = x + \frac{1}{x} = \omega + \frac{1}{\omega} = \omega + \omega^2 = -1
     \]
   - For \( x^2 - 3x + 1 = 0 \):
     Solving the quadratic equation, we get:
     \[
     x = \frac{3 \pm \sqrt{5}}{2}
     \]
     Let \( y = x + \frac{1}{x} \). Then:
     \[
     y = 3 \implies y^2 = 9 \implies x^2 + \frac{1}{x^2} = 7
     \]
     \[
     y^4 = (x^2 + \frac{1}{x^2})^2 = 49 \implies x^4 + \frac{1}{x^4} = 47
     \]
     \[
     y^8 = (x^4 + \frac{1}{x^4})^2 = 2209 \implies x^8 + \frac{1}{x^8} = 2207
     \]
     \[
     y^{16} = (x^8 + \frac{1}{x^8})^2 = 4870849 \implies x^{16} + \frac{1}{x^{16}} = 4870847
     \]

3. **Determine the maximum value:**
   The maximum value of \( x^{16} + \frac{1}{x^{16}} \) is \( 4870847 \).

4. **Find the remainder when \( m \) is divided by 2007:**
   \[
   4870847 \mod 2007
   \]
   We perform the division:
   \[
   4870847 \div 2007 \approx 2427.000497 \implies 4870847 = 2007 \times 2427 + r
   \]
   \[
   r = 4870847 - 2007 \times 2427 = 4870847 - 4870849 = -2
   \]
   Since we need a positive remainder:
   \[
   -2 + 2007 = 2005
   \]

The final answer is \(\boxed{2005}\).