1. Understanding the problem:
We need to find all integer sequences $x_i$ for $1 \le i \le 1997$ that satisfy the equation:
$$\sum_{k=1}^{1997} 2^{k-1} x_k^{1997} = 1996 \prod_{k=1}^{1997} x_k$$

2. Using the hint:
The hint suggests that if $(x_1, x_2, \ldots, x_{1997})$ is a solution, then $\left(\frac{x_1}{2}, \frac{x_2}{2}, \ldots, \frac{x_{1997}}{2}\right)$ is also a solution. This implies that the sequence can be scaled by a factor of $\frac{1}{2}$ and still satisfy the equation.

3. Exploring the implications of the hint:
Let's assume that $(x_1, x_2, \ldots, x_{1997})$ is a solution. Then, substituting $\left(\frac{x_1}{2}, \frac{x_2}{2}, \ldots, \frac{x_{1997}}{2}\right)$ into the equation, we get:
$$\sum_{k=1}^{1997} 2^{k-1} \left(\frac{x_k}{2}\right)^{1997} = 1996 \prod_{k=1}^{1997} \left(\frac{x_k}{2}\right)$$
Simplifying the left-hand side:
$$\sum_{k=1}^{1997} 2^{k-1} \left(\frac{x_k}{2}\right)^{1997} = \sum_{k=1}^{1997} 2^{k-1} \frac{x_k^{1997}}{2^{1997}} = \sum_{k=1}^{1997} \frac{2^{k-1} x_k^{1997}}{2^{1997}}$$
$$= \frac{1}{2^{1997}} \sum_{k=1}^{1997} 2^{k-1} x_k^{1997}$$
Simplifying the right-hand side:
$$1996 \prod_{k=1}^{1997} \left(\frac{x_k}{2}\right) = 1996 \frac{\prod_{k=1}^{1997} x_k}{2^{1997}}$$

4. Equating both sides:
$$\frac{1}{2^{1997}} \sum_{k=1}^{1997} 2^{k-1} x_k^{1997} = 1996 \frac{\prod_{k=1}^{1997} x_k}{2^{1997}}$$
Multiplying both sides by $2^{1997}$:
$$\sum_{k=1}^{1997} 2^{k-1} x_k^{1997} = 1996 \prod_{k=1}^{1997} x_k$$
This is the original equation, confirming that the hint is valid.

5. Considering integer solutions:
Since the sequence can be scaled by $\frac{1}{2}$ and still be a solution, and we are looking for integer solutions, the only possibility is that all $x_i$ must be zero. This is because any non-zero integer scaled by $\frac{1}{2}$ would not remain an integer.

6. Verifying the solution:
If $x_i = 0$ for all $i$, then:
$$\sum_{k=1}^{1997} 2^{k-1} x_k^{1997} = \sum_{k=1}^{1997} 2^{k-1} \cdot 0 = 0$$
$$1996 \prod_{k=1}^{1997} x_k = 1996 \cdot 0 = 0$$
Both sides of the equation are equal, confirming that $x_i = 0$ for all $i$ is indeed a solution.

The final answer is $\boxed{ x_i = 0 }$ for all $i$.