Solve the equation $[x] \cdot \{x\} = 1991x$. (Here $[x]$ denotes the greatest integer less than or equal to $x$, and $\{x\}=x-[x]$.)
Given the equation \([x] \cdot \{x\} = 1991x\), where \([x]\) denotes the greatest integer less than or equal to \(x\), and \(\{x\} = x - [x]\).

1. Let \(a = \lfloor x \rfloor\) and \(b = \{x\}\). Then, we can rewrite the equation as:
   \[
   a \cdot b = 1991(a + b)
   \]

2. Rearrange the equation to isolate terms involving \(a\) and \(b\):
   \[
   ab = 1991a + 1991b
   \]
   \[
   ab - 1991a - 1991b = 0
   \]
   \[
   ab - 1991a - 1991b + 1991^2 = 1991^2
   \]
   \[
   (a - 1991)(b - 1991) = 1991^2
   \]

3. Solve for \(b\) in terms of \(a\):
   \[
   b - 1991 = \frac{1991^2}{a - 1991}
   \]
   \[
   b = 1991 + \frac{1991^2}{a - 1991}
   \]

4. Since \(b = \{x\}\) must be between 0 and 1, we have:
   \[
   0 \leq 1991 + \frac{1991^2}{a - 1991} < 1
   \]

5. Simplify the inequality:
   \[
   0 \leq 1991 + \frac{1991^2}{a - 1991} < 1
   \]
   \[
   -1991 \leq \frac{1991^2}{a - 1991} < -1990
   \]

6. Solve the inequality for \(a\):
   \[
   -1991 \leq \frac{1991^2}{a - 1991} < -1990
   \]
   \[
   \frac{1991^2}{-1991} \geq a - 1991 > \frac{1991^2}{-1990}
   \]
   \[
   -1991 \geq a - 1991 > -1991.0005
   \]
   \[
   0 \geq a > -1.0005
   \]

7. Since \(a\) must be an integer, the possible values for \(a\) are 0 and -1.

8. Substitute \(a = 0\) and \(a = -1\) back into the equation for \(b\):
   - If \(a = 0\):
     \[
     b = 1991 + \frac{1991^2}{0 - 1991} = 1991 - 1991 = 0
     \]
   - If \(a = -1\):
     \[
     b = 1991 + \frac{1991^2}{-1 - 1991} = 1991 - \frac{1991^2}{1992} = 1991 - 1991 = \frac{1991}{1992}
     \]

9. Therefore, the possible values for \(x\) are:
   - If \(a = 0\) and \(b = 0\), then \(x = 0\).
   - If \(a = -1\) and \(b = \frac{1991}{1992}\), then \(x = -1 + \frac{1991}{1992} = -\frac{1}{1992}\).

The final answer is \(\boxed{0 \text{ or } -\frac{1}{1992}}\).