Let $[ {x} ]$ be the greatest integer less than or equal to $x$, and let $\{x\}=x-[x]$. 
Solve the equation: $[x] \cdot \{x\} = 2005x$
1. Given the equation \([x] \cdot \{x\} = 2005x\), we start by expressing \(x\) in terms of its integer and fractional parts:
   \[
   x = [x] + \{x\}
   \]
   where \([x]\) is the greatest integer less than or equal to \(x\) and \(\{x\} = x - [x]\) is the fractional part of \(x\).

2. Substitute \(x = [x] + \{x\}\) into the given equation:
   \[
   [x] \cdot \{x\} = 2005([x] + \{x\})
   \]

3. Distribute the 2005 on the right-hand side:
   \[
   [x] \cdot \{x\} = 2005[x] + 2005\{x\}
   \]

4. Rearrange the equation to isolate \(\{x\}\):
   \[
   [x] \cdot \{x\} - 2005\{x\} = 2005[x]
   \]
   \[
   \{x\}([x] - 2005) = 2005[x]
   \]

5. Solve for \(\{x\}\):
   \[
   \{x\} = \frac{2005[x]}{[x] - 2005}
   \]

6. The condition \(0 \leq \{x\} < 1\) must hold. We analyze this condition for different ranges of \([x]\).

7. Case 1: \([x] < 2005\)
   - If \([x] < 2005\), then \([x] - 2005 < 0\), making the denominator negative.
   - For \(\{x\}\) to be non-negative, the numerator must also be non-positive:
     \[
     2005[x] \leq 0 \implies [x] \leq 0
     \]
   - Let \(n = -[x]\), where \(n \geq 0\). Then:
     \[
     \{x\} = \frac{2005(-n)}{-n - 2005} = \frac{-2005n}{-n - 2005} = \frac{2005n}{n + 2005}
     \]
   - The condition \(0 \leq \{x\} < 1\) becomes:
     \[
     0 \leq \frac{2005n}{n + 2005} < 1
     \]
   - Simplify the inequality:
     \[
     2005n < n + 2005 \implies 2004n < 2005 \implies n < \frac{2005}{2004}
     \]
   - Since \(n\) is an integer, the only possible values are \(n = 0\) or \(n = 1\):
     - If \(n = 0\), then \([x] = 0\) and \(\{x\} = 0\), so \(x = 0\).
     - If \(n = 1\), then \([x] = -1\) and:
       \[
       \{x\} = \frac{2005 \cdot 1}{1 + 2005} = \frac{2005}{2006}
       \]
       \[
       x = [x] + \{x\} = -1 + \frac{2005}{2006} = \frac{-1}{2006}
       \]

8. Case 2: \([x] > 2005\)
   - If \([x] > 2005\), then \([x] - 2005 > 0\), making the denominator positive.
   - For \(\{x\}\) to be less than 1, the numerator must be less than the denominator:
     \[
     2005[x] < [x] - 2005 \implies 2004[x] < -2005 \implies [x] < \frac{-2005}{2004}
     \]
   - Since \([x]\) is an integer, there are no integers less than \(\frac{-2005}{2004}\) that are greater than 2005. Thus, there are no solutions in this case.

9. Therefore, the only solutions are:
   \[
   x = 0 \quad \text{and} \quad x = \frac{-1}{2006}
   \]

The final answer is \(\boxed{x = 0 \text{ and } x = \frac{-1}{2006}}\).