Problem 1394

1. Given the set $S = \{1, 2, 3, x\}$ with four distinct real numbers, we need to find the value of $x$ such that the difference between the largest and smallest values of $S$ is equal to the sum of the elements of $S$.

2. The sum of the elements of $S$ is:
$$1 + 2 + 3 + x = 6 + x$$

3. We need to consider different cases for the value of $x$ to determine the largest and smallest values in $S$.

4. **Case 1: $x \geq 3$**
- The largest value in $S$ is $x$ and the smallest value is $1$.
- The difference between the largest and smallest values is:
$$x - 1$$
- According to the problem, this difference should equal the sum of the elements:
$$x - 1 = 6 + x$$
- Solving for $x$:
$$x - 1 = 6 + x \implies -1 = 6 \quad \text{(impossible)}$$
- Therefore, $x \geq 3$ is not a valid case.

5. **Case 2: $3 > x \geq 1$**
- The largest value in $S$ is $3$ and the smallest value is $1$.
- The difference between the largest and smallest values is:
$$3 - 1 = 2$$
- According to the problem, this difference should equal the sum of the elements:
$$2 = 6 + x$$
- Solving for $x$:
$$2 = 6 + x \implies x = 2 - 6 \implies x = -4$$
- Since $-4$ is not in the interval $3 > x \geq 1$, this case is also not valid.

6. **Case 3: $x < 1$**
- The largest value in $S$ is $3$ and the smallest value is $x$.
- The difference between the largest and smallest values is:
$$3 - x$$
- According to the problem, this difference should equal the sum of the elements:
$$3 - x = 6 + x$$
- Solving for $x$:
$$3 - x = 6 + x \implies 3 - 6 = 2x \implies -3 = 2x \implies x = -\frac{3}{2}$$
- Since $-\frac{3}{2} < 1$, this case is valid.

Therefore, the value of $x$ that satisfies the given conditions is $x = -\frac{3}{2}$.

The final answer is $\boxed{-\frac{3}{2}}$