Let $(x,y,z)$ be an ordered triplet of real numbers that satisfies the following system of equations: \begin{align*}x+y^2+z^4&=0,\\y+z^2+x^4&=0,\\z+x^2+y^4&=0.\end{align*} If $m$ is the minimum possible value of $\lfloor x^3+y^3+z^3\rfloor$, find the modulo $2007$ residue of $m$.
To solve the given system of equations:
\[
\begin{align*}
x + y^2 + z^4 &= 0, \\
y + z^2 + x^4 &= 0, \\
z + x^2 + y^4 &= 0,
\end{align*}
\]
we will analyze the equations step by step.

1. **Assume \(x = y = z = 0\):**
   \[
   \begin{align*}
   0 + 0^2 + 0^4 &= 0, \\
   0 + 0^2 + 0^4 &= 0, \\
   0 + 0^2 + 0^4 &= 0.
   \end{align*}
   \]
   This solution satisfies all three equations. Therefore, \((x, y, z) = (0, 0, 0)\) is a solution.

2. **Check if there are any other non-trivial solutions:**
   Suppose \(x, y, z \neq 0\). We can rewrite the equations as:
   \[
   \begin{align*}
   x &= -y^2 - z^4, \\
   y &= -z^2 - x^4, \\
   z &= -x^2 - y^4.
   \end{align*}
   \]
   By substituting \(x = -y^2 - z^4\) into the second equation:
   \[
   y = -z^2 - (-y^2 - z^4)^4.
   \]
   This substitution leads to a complex polynomial equation, which is difficult to solve analytically. Therefore, we will consider the symmetry and possible values of \(x, y, z\).

3. **Consider the symmetry and possible values:**
   If \(x = y = z\), then:
   \[
   x + x^2 + x^4 = 0.
   \]
   This simplifies to:
   \[
   x(1 + x + x^3) = 0.
   \]
   The solutions to this equation are \(x = 0\) or the roots of \(1 + x + x^3 = 0\). The polynomial \(1 + x + x^3 = 0\) has one real root and two complex roots. The real root is approximately \(x \approx -0.879385\).

4. **Calculate \(x^3 + y^3 + z^3\) for the solutions:**
   - For \(x = y = z = 0\):
     \[
     x^3 + y^3 + z^3 = 0.
     \]
   - For \(x = y = z \approx -0.879385\):
     \[
     x^3 + y^3 + z^3 = 3(-0.879385)^3 \approx -2.275.
     \]

5. **Determine the minimum possible value of \(\lfloor x^3 + y^3 + z^3 \rfloor\):**
   The minimum value is \(\lfloor -2.275 \rfloor = -3\).

6. **Find the modulo 2007 residue of \(m\):**
   \[
   -3 \mod 2007 = 2004.
   \]

The final answer is \(\boxed{2004}\).