Let's determine the positive integers $a, b, c, d$ for which it holds that $a b c d - 1 \mid a + b + c + d$.

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Determine the positive integers $a, b, c, d$ for which it holds that $a b c d - 1 \mid a + b + c + d$.
Due to symmetry, we can assume that $a \leq b \leq c \leq d$. According to the number of 1s in the quadruple, there are five possible cases.

1. Case: $a, b, c$, and $d$ are all equal to 1.

In this case, $a b c d-1=0$, so $a+b+c+d=0$ would need to be true, but this is not the case.

2. Case: Three of $a, b, c$, and $d$ are equal to 1. Then, according to our initial assumption, $a=b=c=1$. Substituting this, we get: $d-1 \mid d+3$. Then $d-1 \mid d+3-(d-1)=4$ also holds.

Since $d$ is a positive integer, the following possibilities exist: $d-1=1$, i.e., $d=2$, and this is a solution; $d-1=2$, i.e., $d=3$, which is also a solution; finally, $d-1=4$, i.e., $d=5$, which is also a solution. Since 4 has only these three positive integer divisors, there will be no more solutions here.

From now on, we will use the following two relationships: if $a b c d-1 \mid a+b+c+d$, then $a b c d-1 \leq a+b+c+d$ (this holds for positive integers); and $a+b+c+d \leq 4 d$.

3. Case: Two of $a, b, c$, and $d$ are equal to 1: $a=b=1$. Then $c d-1 \mid c+d+2$. We will further break down this case based on the value of $c$.

If $c=2$, then we can write the following relationship:

$$
\begin{aligned}
2 d-1 & \leq d+4 \\
d & \leq 5
\end{aligned}
$$

Checking the cases $d=2,3,4,5$, we find that $d=2$ and $d=5$ provide solutions.

If $c=3$, then we can write:

$$
\begin{aligned}
3 d-1 & \leq d+5 \\
d & \leq 3
\end{aligned}
$$

Since $c \leq d$, only the case $d=3$ is possible here, which also provides a solution.

If $c>3$, then we can write:

$$
\begin{aligned}
4 d-1 & \leq c d-1 \leq c+d+2 \leq 2 d+2, \quad \text { from which } \\
2 d & \leq 3
\end{aligned}
$$

Since $d>3$, this is not possible.

Thus, we have found all the solutions here.

4. Case: One of $a, b, c$, and $d$ is equal to 1: $a=1$. Then $b c d-1 \mid b+c+d+1$. Since $2 \leq b \leq c \leq d$, we can write the following relationship:

$$
\begin{aligned}
2 \cdot 2 \cdot d-1=4 d-1 & \leq b c d-1 \leq a+b+c+d=b+c+d+1 \leq 3 d+1, \quad \text { from which } \\
4 d-1 & \leq 3 d+1 \\
d & \leq 2
\end{aligned}
$$

Since $d>1$, only the case $d=2$ is possible here, which provides a solution when $b=c=2$.

5. Case: None of $a, b, c$, and $d$ are 1. Since $2 \leq a \leq b \leq c \leq d$, we can write the following relationship:

$$
\begin{aligned}
2 \cdot 2 \cdot 2 \cdot d-1=8 d-1 & \leq a b c d-1 \leq a+b+c+d \leq 4 d, \quad \text { from which } \\
8 d-1 & \leq 4 d \\
4 d & \leq 1
\end{aligned}
$$

This has no solution for $2 \leq d$, so there is no solution here.

In summary, the following solutions were obtained: $a=1, b=1, c=1, d=2 ; a=1, b=1, c=1, d=3 ; a=1, c=1, d=5 ; a=1, b=1, c=2, d=2 ; a=1, b=1, c=2, d=5 ; a=1, b=1, c=3, d=3 ; a=1, b=2, c=2, d=2$.

Kristóf Kálóczi (Budapesti Fazekas M. Gyak. Ált. Isk. és Gimn., 10th grade)

Remarks. 1. Another thought process used by many: assuming $a \leq b \leq c \leq d$, we get $a b c d-1 \leq a+b+c+d \leq 4 d$. From this, $a b c d \leq 4 d+1 \leq 5 d$, i.e., $a b c \leq 5$ follows. From here (also by case analysis), the solutions can be obtained.

2. In several submissions, the following error occurred: based on the definition of divisibility, they wrote $(a b c d-1) k = a+b+c+d$, from which $a b c d k=a+b+c+d+k$, and also, due to symmetry, they assumed $k \geq d \geq c \geq b \geq a$, from which they got the relationship $a b c d k \leq 5 k$. Finally, dividing by $k$, they arrived at the inequality $a b c d \leq 5$. However, not all solutions are obtained by examining this. The error in the thought process is that $k \geq d$ does not necessarily hold.

3. A third solution can be read on our website.