Polynomial $P(x)=c_{2006}x^{2006}+c_{2005}x^{2005}+\ldots+c_1x+c_0$ has roots $r_1,r_2,\ldots,r_{2006}$.  The coefficients satisfy $2i\tfrac{c_i}{c_{2006}-i}=2j\tfrac{c_j}{c_{2006}-j}$ for all pairs of integers $0\le i,j\le2006$.  Given that $\sum_{i\ne j,i=1,j=1}^{2006} \tfrac{r_i}{r_j}=42$, determine $\sum_{i=1}^{2006} (r_1+r_2+\ldots+r_{2006})$.
1. Given the polynomial \( P(x) = c_{2006}x^{2006} + c_{2005}x^{2005} + \ldots + c_1x + c_0 \) with roots \( r_1, r_2, \ldots, r_{2006} \), we need to determine \( \sum_{i=1}^{2006} r_i \).

2. The coefficients satisfy the condition \( 2i \frac{c_i}{c_{2006} - i} = 2j \frac{c_j}{c_{2006} - j} \) for all pairs of integers \( 0 \le i, j \le 2006 \). This implies that the ratio \( \frac{c_i}{c_{2006} - i} \) is constant for all \( i \). Let this constant be \( k \). Therefore, we have:
   \[
   \frac{c_i}{c_{2006} - i} = k \quad \text{for all } i
   \]
   This implies:
   \[
   c_i = k (c_{2006} - i)
   \]

3. Given that \( \sum_{i \ne j, i=1, j=1}^{2006} \frac{r_i}{r_j} = 42 \), we need to use properties of the roots of the polynomial. By Vieta's formulas, the sum of the roots of the polynomial \( P(x) \) is given by:
   \[
   \sum_{i=1}^{2006} r_i = -\frac{c_{2005}}{c_{2006}}
   \]

4. To find \( c_{2005} \), we use the relationship derived from the given condition. For \( i = 2005 \):
   \[
   c_{2005} = k (c_{2006} - 2005)
   \]
   For \( i = 2006 \):
   \[
   c_{2006} = k (c_{2006} - 2006)
   \]
   Solving for \( k \):
   \[
   c_{2006} = k (c_{2006} - 2006) \implies k = \frac{c_{2006}}{c_{2006} - 2006}
   \]

5. Substituting \( k \) back into the expression for \( c_{2005} \):
   \[
   c_{2005} = \frac{c_{2006}}{c_{2006} - 2006} (c_{2006} - 2005)
   \]
   Simplifying:
   \[
   c_{2005} = \frac{c_{2006} (c_{2006} - 2005)}{c_{2006} - 2006}
   \]

6. Using Vieta's formula:
   \[
   \sum_{i=1}^{2006} r_i = -\frac{c_{2005}}{c_{2006}} = -\frac{\frac{c_{2006} (c_{2006} - 2005)}{c_{2006} - 2006}}{c_{2006}} = -\frac{c_{2006} - 2005}{c_{2006} - 2006}
   \]

7. Simplifying further:
   \[
   \sum_{i=1}^{2006} r_i = -\frac{c_{2006} - 2005}{c_{2006} - 2006} = -\left(1 - \frac{1}{c_{2006} - 2006}\right) = -1 + \frac{1}{c_{2006} - 2006}
   \]

8. Given that \( \sum_{i \ne j, i=1, j=1}^{2006} \frac{r_i}{r_j} = 42 \), we can infer that the polynomial has a specific structure. However, the exact value of \( c_{2006} \) is not provided, and thus we cannot simplify further without additional information.

The final answer is \(\boxed{-1 + \frac{1}{c_{2006} - 2006}}\).