Find all quadruples $(a, b, c, d)$ of real numbers such that
\[a + bcd = b + cda = c + dab = d + abc.\]
We are given the system of equations:
\[a + bcd = b + cda = c + dab = d + abc.\]

1. **Introduce a common variable:**
   Let \( Q = a + bcd = b + cda = c + dab = d + abc \).

2. **Rewrite each equation:**
   \[
   a + bcd = Q \implies a = Q - bcd,
   \]
   \[
   b + cda = Q \implies b = Q - cda,
   \]
   \[
   c + dab = Q \implies c = Q - dab,
   \]
   \[
   d + abc = Q \implies d = Q - abc.
   \]

3. **Analyze the polynomial form:**
   Consider the polynomial form of each equation:
   \[
   a + \frac{P}{a} = Q \implies a^2 - Qa + P = 0,
   \]
   where \( P = abcd \). Similarly, we get:
   \[
   b^2 - Qb + P = 0,
   \]
   \[
   c^2 - Qc + P = 0,
   \]
   \[
   d^2 - Qd + P = 0.
   \]

4. **Possible values for \(a, b, c, d\):**
   Since each of these quadratic equations must have the same roots, \(a, b, c, d\) can take at most two different values. Without loss of generality, assume \( \{a, b, c, d\} = \{a, a, b, b\} \).

5. **Solve for \(a\) and \(b\):**
   Substitute into the original equations:
   \[
   a + ab^2 = b + a^2b \implies (a - b)(ab - 1) = 0.
   \]
   This gives us two cases:
   - \(a = b\), which leads to the solution \((a, a, a, a)\).
   - \(ab = 1\), which leads to the solution \(\left(a, a, \frac{1}{a}, \frac{1}{a}\right)\) for \(a \neq 0\).

6. **Consider the case of three equal values:**
   If \( \{a, b, c, d\} = \{a, a, a, b\} \), then:
   \[
   a + a^2b = b + a^3 \implies (a - b)(1 - a^2) = 0.
   \]
   This gives us two cases:
   - \(a = b\), which is already covered.
   - \(a = \pm 1\), leading to the solutions \((1, 1, 1, b)\) and \((-1, -1, -1, b)\).

7. **Consider the case of zero values:**
   If one of the values is zero, we have:
   - \((0, 0, x, y)\), which does not work as shown in the problem.
   - \((0, x, x, x)\), which also does not provide new solutions.

8. **Summarize the solutions:**
   The non-ordered quadruples are:
   - \((a, a, a, a)\) for any \(a \in \mathbb{R}\),
   - \(\left(a, a, \frac{1}{a}, \frac{1}{a}\right)\) for \(a \neq 0\),
   - \((1, 1, 1, b)\) for any \(b \in \mathbb{R}\),
   - \((-1, -1, -1, b)\) for any \(b \in \mathbb{R}\).

The final answer is \( \boxed{ (a, a, a, a) } \), \(\left(a, a, \frac{1}{a}, \frac{1}{a}\right)\) for \(a \neq 0\), \((1, 1, 1, b)\), and \((-1, -1, -1, b)\).