A sequence of real numbers $ x_0, x_1, x_2, \ldots$ is defined as follows: $ x_0 \equal{} 1989$ and for each $ n \geq 1$

\[ x_n \equal{} \minus{} \frac{1989}{n} \sum^{n\minus{}1}_{k\equal{}0} x_k.\]

Calculate the value of $ \sum^{1989}_{n\equal{}0} 2^n x_n.$
1. **Define the sequence and initial conditions:**
   The sequence \( \{x_n\} \) is defined by:
   \[
   x_0 = 1989
   \]
   and for \( n \geq 1 \),
   \[
   x_n = -\frac{1989}{n} \sum_{k=0}^{n-1} x_k.
   \]

2. **Calculate the first few terms:**
   - For \( n = 1 \):
     \[
     x_1 = -\frac{1989}{1} \sum_{k=0}^{0} x_k = -1989 \cdot x_0 = -1989 \cdot 1989 = -1989^2.
     \]
   - For \( n = 2 \):
     \[
     x_2 = -\frac{1989}{2} \sum_{k=0}^{1} x_k = -\frac{1989}{2} (x_0 + x_1) = -\frac{1989}{2} (1989 - 1989^2).
     \]

3. **General form of \( x_n \):**
   We observe that:
   \[
   x_{n+1} = -\frac{1989}{n+1} \left( \sum_{k=0}^{n} x_k \right).
   \]
   Using the recurrence relation, we can derive:
   \[
   x_{n+1} = -\frac{1989}{n+1} \left( x_n + \sum_{k=0}^{n-1} x_k \right).
   \]
   Simplifying further, we get:
   \[
   x_{n+1} = -\frac{1989}{n+1} \left( x_n + \frac{n}{1989} x_n \right) = -\frac{1989}{n+1} \left( \frac{1989 + n}{1989} x_n \right) = -\frac{1989 + n}{n+1} x_n.
   \]

4. **Pattern recognition:**
   We notice that:
   \[
   x_n = (-1)^n 1989 \binom{1989}{n}.
   \]
   This can be verified by induction or by observing the pattern in the sequence.

5. **Sum calculation:**
   We need to calculate:
   \[
   \sum_{n=0}^{1989} 2^n x_n.
   \]
   Substituting the general form of \( x_n \):
   \[
   \sum_{n=0}^{1989} 2^n x_n = \sum_{n=0}^{1989} 2^n (-1)^n 1989 \binom{1989}{n}.
   \]
   Factor out the constant:
   \[
   1989 \sum_{n=0}^{1989} (-2)^n \binom{1989}{n}.
   \]
   Using the binomial theorem, we know:
   \[
   \sum_{n=0}^{1989} \binom{1989}{n} (-2)^n = (1 - 2)^{1989} = (-1)^{1989} = -1.
   \]
   Therefore:
   \[
   1989 \sum_{n=0}^{1989} (-2)^n \binom{1989}{n} = 1989 \cdot (-1) = -1989.
   \]

The final answer is \(\boxed{-1989}\).