Find the lowest possible values from the function
\[ f(x) \equal{} x^{2008} \minus{} 2x^{2007} \plus{} 3x^{2006} \minus{} 4x^{2005} \plus{} 5x^{2004} \minus{} \cdots \minus{} 2006x^3 \plus{} 2007x^2 \minus{} 2008x \plus{} 2009\]
for any real numbers $ x$.
To find the lowest possible value of the function
\[ f(x) = x^{2008} - 2x^{2007} + 3x^{2006} - 4x^{2005} + \cdots - 2006x^3 + 2007x^2 - 2008x + 2009, \]
we will use induction and properties of polynomials.

1. **Base Case:**
   Consider the polynomial for \( n = 1 \):
   \[ f_2(x) = x^2 - 2x + 3. \]
   We can rewrite this as:
   \[ f_2(x) = (x-1)^2 + 2. \]
   Clearly, the minimum value of \( f_2(x) \) occurs at \( x = 1 \), and the minimum value is:
   \[ f_2(1) = 2. \]

2. **Induction Hypothesis:**
   Assume that for some \( n \), the polynomial
   \[ f_{2n}(x) = x^{2n} - 2x^{2n-1} + 3x^{2n-2} - \cdots - 2nx + (2n+1) \]
   can be written as:
   \[ f_{2n}(x) = (x-1)^2(x^{2n-2} + 2x^{2n-3} + \cdots + n) + (n+1). \]
   This implies that the minimum value of \( f_{2n}(x) \) is \( n+1 \) and occurs at \( x = 1 \).

3. **Induction Step:**
   We need to show that the polynomial for \( n+1 \):
   \[ f_{2n+2}(x) = x^{2n+2} - 2x^{2n+1} + 3x^{2n} - \cdots - (2n+2)x + (2n+3) \]
   can be written in a similar form. Using the induction hypothesis:
   \[ f_{2n+2}(x) = x^2 f_{2n}(x) - (2n+2)x + (2n+3). \]
   Substituting the expression for \( f_{2n}(x) \):
   \[ f_{2n+2}(x) = x^2 \left( (x-1)^2(x^{2n-2} + 2x^{2n-3} + \cdots + n) + (n+1) \right) - (2n+2)x + (2n+3). \]
   Simplifying, we get:
   \[ f_{2n+2}(x) = (x-1)^2(x^{2n} + 2x^{2n-1} + \cdots + nx^2) + (n+1)(x-1)^2 + (2n+3). \]
   \[ f_{2n+2}(x) = (x-1)^2(x^{2n} + 2x^{2n-1} + \cdots + nx^2 + n+1) + (2n+3). \]
   Therefore, the minimum value of \( f_{2n+2}(x) \) is \( n+2 \) and occurs at \( x = 1 \).

4. **Conclusion:**
   By induction, the minimum value of \( f_{2008}(x) \) is \( 1005 \) and occurs at \( x = 1 \).

The final answer is \( \boxed{1005} \).