Find all sets of four real numbers $x_1, x_2, x_3, x_4$ such that the sum of any one and the product of the other three is equal to 2.
1. **Given Equations:**
   We start with the given equations:
   \[
   x_1 + x_2 x_3 x_4 = 2 \quad (1)
   \]
   \[
   x_2 + x_3 x_4 x_1 = 2 \quad (2)
   \]
   \[
   x_3 + x_4 x_1 x_2 = 2 \quad (3)
   \]
   \[
   x_4 + x_1 x_2 x_3 = 2 \quad (4)
   \]

2. **Subtracting Equations:**
   Subtract equation (2) from equation (1):
   \[
   (x_1 + x_2 x_3 x_4) - (x_2 + x_3 x_4 x_1) = 2 - 2
   \]
   Simplifying, we get:
   \[
   x_1 - x_2 + x_2 x_3 x_4 - x_3 x_4 x_1 = 0
   \]
   Factor out \((x_1 - x_2)\):
   \[
   (x_1 - x_2) - x_3 x_4 (x_1 - x_2) = 0
   \]
   \[
   (x_1 - x_2)(1 - x_3 x_4) = 0
   \]
   This implies:
   \[
   x_1 = x_2 \quad \text{or} \quad x_3 x_4 = 1
   \]

3. **General Case:**
   By symmetry, subtracting any pair of equations \(a\) and \(b\) (where \(1 \leq a \neq b \leq 4\)) will yield:
   \[
   x_a = x_b \quad \text{or} \quad x_c x_d = 1
   \]
   where \(\{a, b, c, d\} = \{1, 2, 3, 4\}\).

4. **Considering Equalities:**
   - **Case 1: All four numbers are equal.**
     If \(x_1 = x_2 = x_3 = x_4 = x\), substituting into any equation:
     \[
     x + x^3 = 2
     \]
     \[
     x^3 + x - 2 = 0
     \]
     The only real solution to this cubic equation is \(x = 1\). Thus:
     \[
     (1, 1, 1, 1)
     \]

   - **Case 2: Three numbers are equal, and one is distinct.**
     Without loss of generality, let \(x_1 = x_2 = x_3\) and \(x_4\) be distinct. Then:
     \[
     x_1 + x_1^2 x_4 = 2
     \]
     \[
     x_1^2 x_4 = 2 - x_1
     \]
     Also, since \(x_1 x_1 = 1\):
     \[
     x_1^2 = 1 \implies x_1 = \pm 1
     \]
     - If \(x_1 = 1\):
       \[
       1 + 1 \cdot x_4 = 2 \implies x_4 = 1
       \]
       This gives \((1, 1, 1, 1)\), which is already considered.
     - If \(x_1 = -1\):
       \[
       -1 + (-1)^2 x_4 = 2 \implies -1 + x_4 = 2 \implies x_4 = 3
       \]
       This gives \((-1, -1, -1, 3)\).

5. **Permutations:**
   The solution \((-1, -1, -1, 3)\) can be permuted to give all possible sets:
   \[
   (-1, -1, -1, 3), (-1, -1, 3, -1), (-1, 3, -1, -1), (3, -1, -1, -1)
   \]

The final answer is \( \boxed{ (1, 1, 1, 1) } \) and \((-1, -1, -1, 3)\) and its permutations.