It is known that $a^{2005} + b^{2005}$ can be expressed as the polynomial of $a + b$ and $ab$. Find the coefficients' sum of this polynomial.
1. **Expressing \(a^{2005} + b^{2005}\) as a polynomial \(P(a+b, ab)\):**
   Given that \(a^{2005} + b^{2005}\) can be expressed as a polynomial \(P(a+b, ab)\), we need to find the sum of the coefficients of this polynomial. The sum of the coefficients of a polynomial \(P(x, y)\) is given by evaluating the polynomial at \(x = 1\) and \(y = 1\).

2. **Choosing specific values for \(a\) and \(b\):**
   Let \(a = \frac{1 + i\sqrt{3}}{2}\) and \(b = \frac{1 - i\sqrt{3}}{2}\). These values are chosen because they are the cube roots of unity, which have useful properties:
   \[
   a + b = \left(\frac{1 + i\sqrt{3}}{2}\right) + \left(\frac{1 - i\sqrt{3}}{2}\right) = 1
   \]
   \[
   ab = \left(\frac{1 + i\sqrt{3}}{2}\right) \left(\frac{1 - i\sqrt{3}}{2}\right) = \frac{1 - (i\sqrt{3})^2}{4} = \frac{1 + 3}{4} = 1
   \]

3. **Evaluating \(P(1, 1)\):**
   Since \(a\) and \(b\) are cube roots of unity, we know that:
   \[
   a^3 = 1 \quad \text{and} \quad b^3 = 1
   \]
   Therefore, \(a^{2005}\) and \(b^{2005}\) can be simplified using the properties of exponents:
   \[
   a^{2005} = (a^3)^{668} \cdot a = 1^{668} \cdot a = a
   \]
   \[
   b^{2005} = (b^3)^{668} \cdot b = 1^{668} \cdot b = b
   \]
   Thus,
   \[
   a^{2005} + b^{2005} = a + b = 1
   \]

4. **Conclusion:**
   Since \(P(a+b, ab) = P(1, 1)\) and we have shown that \(a^{2005} + b^{2005} = 1\), it follows that:
   \[
   P(1, 1) = 1
   \]

The final answer is \(\boxed{1}\).