Determine all four-digit numbers $\overline{a b c d}$ such that

$$
(a+b)(a+c)(a+d)(b+c)(b+d)(c+d)=\overline{a b c d}
$$
Depending on the parity of $a, b, c, d$, at least two of the factors $(a+b),(a+c)$, $(a+d),(b+c),(b+d),(c+d)$ are even, so that $4 \mid \overline{a b c d}$.

We claim that $3 \mid \overline{a b c d}$.

Assume $a+b+c+d \equiv 2(\bmod 3)$. Then $x+y \equiv 1(\bmod 3)$, for all distinct $x, y \in\{a, b, c, d\}$. But then the left hand side in the above equality is congruent to $1(\bmod 3)$ and the right hand side congruent to $2(\bmod 3)$, contradiction.

Assume $a+b+c+d \equiv 1(\bmod 3)$. Then $x+y \equiv 2(\bmod 3)$, for all distinct $x, y \in\{a, b, c, d\}$, and $x \equiv 1(\bmod 3)$, for all $x, y \in\{a, b, c, d\}$. Hence, $a, b, c, d \in\{1,4,7\}$, and since $4 \mid \overline{a b c d}$, we have $c=d=4$. Therefore, $8 \mid \overline{a b 44}$, and since at least one more factor is even, it follows that $16 \mid \overline{a b 44}$. Then $b \neq 4$, and the only possibilities are $b=1$, implying $a=4$, which is impossible because 4144 is not divisible by $5=1+4$, or $b=7$, implying $11 \mid \overline{a 744}$, hence $a=7$, which is also impossible because 7744 is not divisible by $14=7+7$.

We conclude that $3 \mid \overline{a b c d}$, hence also $3 \mid a+b+c+d$. Then at least one factor $x+y$ of $(a+b),(a+c),(a+d),(b+c),(b+d),(c+d)$ is a multiple of 3, implying that also $3 \mid a+b+c+d-x-y$, so $9 \mid \overline{a b c d}$. Then $9 \mid a+b+c+d$, and $a+b+c+d \in\{9,18,27,36\}$. Using the inequality $x y \geq x+y-1$, valid for all $x, y \in \mathbb{N}^{*}$, if $a+b+c+d \in\{27,36\}$, then

$$
\overline{a b c d}=(a+b)(a+c)(a+d)(b+c)(b+d)(c+d) \geq 26^{3}>10^{4}
$$

which is impossible.

Using the inequality $x y \geq 2(x+y)-4$ for all $x, y \geq 2$, if $a+b+c+d=18$ and all two-digit sums are greater than 1, then $\overline{a b c d} \geq 32^{3}>10^{4}$. Hence, if $a+b+c+d=18$, some two-digit sum must be 1, hence the complementary sum will be 17, and the digits are $\{a, b, c, d\}=\{0,1,8,9\}$. But then $\overline{a b c d}=1 \cdot 17 \cdot 8 \cdot 9^{2} \cdot 10>10^{4}$.

We conclude that $a+b+c+d=9$. Then among $a, b, c, d$ there are either three odd or three even numbers, and $8 \mid \overline{a b c d}$.

If three of the digits are odd, then $d$ is even and since $c$ is odd, divisibility by 8 implies that $d \in\{2,6\}$. If $d=6$, then $a=b=c=1$. But 1116 is not divisible by 7, so this is not a solution. If $d=2$, then $a, b, c$ are either $1,1,5$ or $1,3,3$ in some order. In the first case $2 \cdot 6^{2} \cdot 3^{2} \cdot 7=4536 \neq \overline{a b c d}$. The second case cannot hold because the resulting number is not a multiple of 5.

Hence, there has to be one odd and three even digits. At least one of the two-digit sums of even digits is a multiple of 4, and since there cannot be two zero digits, we have either $x+y=4$ and $z+t=5$, or $x+y=8$ and $z+t=1$ for some ordering $x, y, z, t$ of $a, b, c, d$. In the first case we have $d=0$ and the digits are $0,1,4,4$, or $0,2,3,4$, or $0,2,2,5$. None of these is a solution because $1 \cdot 4^{2} \cdot 5^{2} \cdot 8=3200,2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7=5040$ and $2^{2} \cdot 5 \cdot 4 \cdot 7^{2}=3920$. In the second case two of the digits are 0 and 1, and the other two have to be either 4 and 4, or 2 and 6. We already know that the first possibility fails. For the second, we get

$$
(0+1) \cdot(0+2) \cdot(0+6) \cdot(1+2) \cdot(1+6) \cdot(2+6)=2016
$$

and $\overline{a b c d}=2016$ is the only solution.