Let $c$ be a fixed real number. Show that a root of the equation
\[x(x+1)(x+2)\cdots(x+2009)=c\]
can have multiplicity at most $2$. Determine the number of values of $c$ for which the equation has a root of multiplicity $2$.
1. **Multiplicity of Roots:**

   To show that a root of the equation 
   \[
   x(x+1)(x+2)\cdots(x+2009) = c
   \]
   can have multiplicity at most 2, we start by defining the function 
   \[
   f(x) = \prod_{k=0}^{2009} (x+k) - c.
   \]
   The derivative of \( f(x) \) is given by:
   \[
   f'(x) = \prod_{k=0}^{2009} (x+k) \sum_{k=0}^{2009} \frac{1}{x+k}.
   \]
   The second derivative of \( f(x) \) is:
   \[
   f''(x) = f'(x) \sum_{k=0}^{2009} \frac{1}{x+k} - \prod_{k=0}^{2009} (x+k) \sum_{k=0}^{2009} \frac{1}{(x+k)^2}.
   \]
   If \( r \) is a root of multiplicity greater than 2, then \( f(r) = f'(r) = f''(r) = 0 \). This implies:
   \[
   \prod_{k=0}^{2009} (r+k) \sum_{k=0}^{2009} \frac{1}{(r+k)^2} = 0.
   \]
   Since \( \prod_{k=0}^{2009} (r+k) = c \neq 0 \), it follows that:
   \[
   \sum_{k=0}^{2009} \frac{1}{(r+k)^2} = 0,
   \]
   which is impossible because the sum of positive terms cannot be zero. Therefore, a root of the equation can have multiplicity at most 2.

2. **Number of Values of \( c \) for Double Roots:**

   The number of double roots of \( f(x) \) is the number of roots of \( f'(x) \). We need to determine the number of distinct values of \( c \) for which \( f(x) \) has a double root.

   The function \( f'(x) \) is:
   \[
   f'(x) = \prod_{k=0}^{2009} (x+k) \sum_{k=0}^{2009} \frac{1}{x+k}.
   \]
   The roots of \( f'(x) \) are the values of \( x \) for which:
   \[
   \sum_{k=0}^{2009} \frac{1}{x+k} = 0.
   \]
   This is a rational function with \( 2009 \) simple poles at \( x = -k \) for \( k = 0, 1, \ldots, 2009 \). By the intermediate value theorem, there are \( 2009 \) real roots of \( f'(x) \).

   Each root \( r \) of \( f'(x) \) corresponds to a value of \( c \) given by:
   \[
   c = \prod_{k=0}^{2009} (r+k).
   \]
   For each root \( r \), there is a corresponding root \( r' = -2009 - r \) such that:
   \[
   \prod_{k=0}^{2009} (r+k) = \prod_{k=0}^{2009} (-2009-r+k).
   \]
   This implies that each pair \( (r, -2009-r) \) gives the same value of \( c \). Since there are \( 2009 \) roots of \( f'(x) \), there are \( \frac{2009}{2} = 1004.5 \) pairs, which rounds to \( 1005 \) distinct values of \( c \).

   Therefore, the number of values of \( c \) for which the equation has a root of multiplicity 2 is \( \boxed{1005} \).