18. (2004 Australian Mathematical Olympiad) Find all quadruples of real numbers $(a, b, c, d)$ such that:
the sum of the product of any three of the numbers and the remaining one is the same, regardless of the choice of the three numbers.
18. The given problem is to find all real quadruples \((a, b, c, d)\) that satisfy the system of equations
$$
a b c+d=a b d+c=a c d+b=b c d+a
$$

Since the system (1) is symmetric with respect to all variables, any permutation of a solution is also a solution. Noting this, we can consider the following 5 cases:
(1) \(a=b=c=d\);
(2) \(a=b=c \neq d\);
(3) \(a=b \neq c=d\);
(4) \(a=b \neq c\) and \(c \neq d, d \neq b\);
(5) No two are equal.
From the first equation of (1), \(a b c+d=a b d+c\), we get \(a b c-a b d-c+d=0\), which simplifies to \((c-d)(a b-1)=0\).
Therefore, from the given system (1), we can also derive similar equations:
$$
\begin{array}{l}
(a-b)(c d-1)=0, \\
(a-c)(b d-1)=0, \\
(a-d)(b c-1)=0, \\
(b-c)(a d-1)=0, \\
(b-d)(a c-1)=0 .
\end{array}
$$

Now, let's consider each possible case.
(1) If \(a=b=c=d=r\), then for any real number \(r\), it is a solution. In this case, all 4 expressions in (1) equal \(r^{3}+r\), so the solution set is \(S_{1}=\{(r, r, r, r) \mid r \in \mathbf{R}\}\).
(2) If \(a=b=c \neq d\), from \((c-d)(a b-1)=0\), we get \(a b=1\), which means \(b=\frac{1}{a}\) must hold.
Since \(a=b\), we have \(a^{2}=1\), hence \(a= \pm 1\).
In fact, taking \(a=b=c= \pm 1\) and \(d=r \neq a\), all 4 expressions in (1) equal \(r \pm 1\). Considering all permutations, the solution set for this case is
$$
S_{2}=\{( \pm 1, \pm 1, \pm 1, r),( \pm 1, \pm 1, r, \pm 1),( \pm 1, r, \pm 1, \pm 1),(r, \pm 1, \pm 1, \pm 1) \mid r \in \mathbf{R}\} .
$$
(3) If \(a=b \neq c=d\), from \((b-c)(a d-1)=0\), we get \(a d=1\), which means \(d=\frac{1}{a}\).
In fact, taking \(a=b=r\) and \(c=d=\frac{1}{r}\), all 4 expressions in (1) equal \(r+\frac{1}{r}\). Considering all permutations, the solution set for this case is
$$
S_{3}=\left\{\left(r, r, \frac{1}{r}, \frac{1}{r}\right),\left(r, \frac{1}{r}, r, \frac{1}{r}\right), \left.\left(r, \frac{1}{r}, \frac{1}{r}, r\right) \right\rvert\, r \in \mathbf{R}, r \neq 0\right\} .
$$
(4) If \(a=b \neq c\) and \(c \neq d, d \neq b\), we get
\(1=a d=a b=a c\), which implies \(b=c=d\), a contradiction.
Therefore, there is no solution in this case.
(5) For the same reason as (4), there is no solution in this case.
In summary, the set of all solutions to the given system of equations is
$$
S=S_{1} \cup S_{2} \cup S_{3} .
$$