4・184 For any three positive integers $x, y, z$, let
$$f(x, y, z)=[1+2+3+\cdots+(x+y-2)]-z$$

Find all positive integer quadruples $(a, b, c, d)$, such that
$$f(a, b, c)=f(c, d, a)=1993$$
[Solution] Let $(a, b, c, d)$ be the required positive integer tuple. Then, by the problem statement, we have
$$\left\{\begin{array}{l}
f(a, b, c)=1993 \\
f(c, d, a)=1993
\end{array}\right.$$

Since
$$f(x, y, z)=\frac{1}{2}(x+y-2)(x+y-1)-z,$$

we have
$$\left\{\begin{array}{l}
\frac{1}{2}(a+b-2)(a+b-1)-c=1993 \\
\frac{1}{2}(c+d-2)(c+d-1)-a=1993 .
\end{array}\right.$$

From the above two equations, we get
$$(a+b-2)(a+b-1)-2 c=(c+d-2)(c+d-1)-2 a,$$

which can be rewritten as
$$(a+b-1)^{2}+(a-b+1)=(c+d-1)^{2}+(c-d+1)$$

Rearranging terms, we get
$$(a+b-1)^{2}-(c+d-1)^{2}=(c-d+1)-(a-b+1)$$

which simplifies to
$$\begin{array}{l}
(a+b+c+d-2)(a+b-c-d)=(c-a)+(b-d), \\
M=a+b+c+d, \quad M \geqslant 4 .
\end{array}$$

Rewriting the above equation, we get
$$(a-c)(M-1)=(d-b)(M-3)$$

We now prove that $d=b$.
By contradiction. Suppose $d \neq b$, without loss of generality, assume $b < d$. (The case $b > d$ can be similarly proven.) From (2), we get
$$a > c .$$

Since
$$(M-1)-(M-3)=2$$

the greatest common divisor of $M-1$ and $M-3$ can only be 1 or 2.
If $(M-1, M-3)=1$, then from (2) we know that $d-b$ is a multiple of $M-1$, which contradicts
$$M-1=a+b+c+d-1>b+c+d>d-b$$

Therefore,
$$(M-1, M-3)=2 .$$

Thus, $M-1$ and $M-3$ are both even, and $\frac{1}{2}(M-1)$ and $\frac{1}{2}(M-3)$ are coprime. From (2), we get
$$(a-c) \cdot \frac{1}{2}(M-1)=(d-b) \cdot \frac{1}{2}(M-3)$$

This implies that $d-b$ is a multiple of $\frac{1}{2}(M-1)$, and $a-c$ is a multiple of $\frac{1}{2}(M-3)$, so $4(a-c)(d-b)$ is a multiple of $(M-1)(M-3)$. However,
$$\begin{aligned}
& (M-1)(M-3)-4(a-c)(d-b) \\
= & \left(M^{2}-4 M+3\right)-4(a d-c d-a b+b c) \\
= & (a+b+c+d)^{2}-4(a+b+c+d)+3-4(a d-c d-a b \\
& +b c) \\
= & \left(a^{2}+b^{2}+c^{2}+d^{2}+2 a c+2 b d-2 a b-2 a d-2 b c-2 c d\right) \\
& +8 a b+8 c d-4(a+b+c+d)+3 \\
= & (a-b+c-d)^{2}+4[a(b-1)+b(a-1)+c(d-1) \\
& +d(c-1)]+3
\end{aligned}$$
$>0$, which leads to a contradiction.
Therefore, we must have $b=d$, and from (2) we get $a=c$.
Substituting this result into (1), we get
$$\frac{1}{2}(a+b-2)(a+b-1)-a=1993,$$

which simplifies to
$$(a+b-2)(a+b-1)=3986+2 a,$$

Noting that $a+b-2$ and $a+b-1$ are two consecutive natural numbers, and $62 \times 63=$
$3906,63 \times 64=4032$, we have
$$a+b-2 \geqslant 63 .$$

On the other hand, from (3), we get
$$(a+b-3)(a+b-2)=3986+2 a-2(a+b-2)$$

which simplifies to
$$(a+b-3)(a+b-2)=3990-2 b,$$

Clearly,
$$(a+b-3)(a+b-2)<3990$$

Noting that $62 \times 63=3906,63 \times 64=4032$, we have
$$a+b-2 \leqslant 63$$

From (4) and (5), we get
$$a+b-2=63,$$

which simplifies to
$$a+b=65 .$$

Substituting this result into (3), we get
$$63 \times 64=3986+2 a,$$

Thus, we have
$$a=23, b=42(a=c, b=d) .$$

Therefore, the only positive integer tuple that satisfies the problem is $(23,42,23,42)$.