Given that for reals $a_1,\cdots, a_{2004},$ equation $x^{2006}-2006x^{2005}+a_{2004}x^{2004}+\cdots +a_2x^2+a_1x+1=0$ has $2006$ positive real solution, find the maximum possible value of $a_1.$
1. **Using Vieta's Formulas:**
   Given the polynomial \( P(x) = x^{2006} - 2006x^{2005} + a_{2004}x^{2004} + \cdots + a_2x^2 + a_1x + 1 \), we know from Vieta's formulas that the sum of the roots \( x_1, x_2, \ldots, x_{2006} \) is given by:
   \[
   \sum_{k=1}^{2006} x_k = 2006
   \]
   and the sum of the reciprocals of the roots is:
   \[
   \sum_{k=1}^{2006} \frac{1}{x_k} = -a_1
   \]

2. **Applying the AM-HM Inequality:**
   The Arithmetic Mean-Harmonic Mean (AM-HM) inequality states that for any set of positive real numbers \( x_1, x_2, \ldots, x_n \):
   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n}}
   \]
   Applying this to our set of roots \( x_1, x_2, \ldots, x_{2006} \), we get:
   \[
   \frac{\sum_{k=1}^{2006} x_k}{2006} \geq \frac{2006}{\sum_{k=1}^{2006} \frac{1}{x_k}}
   \]
   Substituting the known sums from Vieta's formulas:
   \[
   \frac{2006}{2006} \geq \frac{2006}{-a_1}
   \]
   Simplifying, we get:
   \[
   1 \geq \frac{2006}{-a_1}
   \]
   Multiplying both sides by \(-a_1\) (noting that \(-a_1\) is positive since the roots are positive):
   \[
   -a_1 \geq 2006
   \]
   Therefore:
   \[
   a_1 \leq -2006
   \]

3. **Conclusion:**
   The maximum possible value of \( a_1 \) is \(-2006\).

The final answer is \(\boxed{-2006}\)