Junior Balkan Mathematics Olympiad

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