For which pairwise distinct positive integers $a, b, c, d$ will the value of

$$
\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}
$$

be an integer?
Solution. Let $x^{\prime}=x+1$. Adding the quantity

$$
S=\frac{1}{a^{\prime}}+\frac{1}{b^{\prime}}+\frac{1}{c^{\prime}}+\frac{1}{d^{\prime}}
$$

to the expression, we get an integer (4). Therefore, a necessary and sufficient condition is that the value of $S$ is also an integer for integers $a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}$ greater than 1 and pairwise distinct. Since

$$
S \leq \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}<2
$$

this can only happen if $S=1$. Let us assume for simplicity that $a^{\prime}<b^{\prime}<c^{\prime}<d^{\prime}$. If $a^{\prime} \geq 3$, then $S \leq$ $\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}<1$, so it must be that $a^{\prime}=2$,

$$
S^{\prime}=\frac{1}{b^{\prime}}+\frac{1}{c^{\prime}}+\frac{1}{d^{\prime}}=\frac{1}{2}
$$

If $b^{\prime} \geq 6$, then $S \leq \frac{1}{2}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}<1$, so $3 \leq b^{\prime} \leq 5$. The case $b^{\prime}=5$ can be easily ruled out, as then $c^{\prime} \geq 6$. If $c^{\prime}=6$, then

$$
\frac{1}{d^{\prime}}=\frac{1}{2}-\frac{1}{5}-\frac{1}{6}=\frac{2}{15}
$$

which is not possible. If $c^{\prime} \geq 7$, then $d^{\prime} \geq 8$, and thus $S^{\prime}<1 / 2$.

In the case $b^{\prime}=4$,

$$
\frac{1}{c^{\prime}}+\frac{1}{d^{\prime}}=\frac{1}{4}
$$

or $c^{\prime} d^{\prime}=4 c^{\prime}+4 d^{\prime}$, and thus $\left(c^{\prime}-4\right)\left(d^{\prime}-4\right)=16$. Since $1 \leq c^{\prime}-4<d^{\prime}-4$ are integers, this is only possible if $c^{\prime}-4=1$ and $d^{\prime}-4=16$, or $c^{\prime}-4=2, d^{\prime}-4=8$. In the first case, $c^{\prime}=5$ and

$d^{\prime}=20$, in the second case, $c^{\prime}=6$ and $d^{\prime}=12$. In the case $b^{\prime}=3$, a similar line of reasoning from $\frac{1}{c^{\prime}}+\frac{1}{d^{\prime}}=\frac{1}{6}$ gives $\left(c^{\prime}-6\right)\left(d^{\prime}-6\right)=36$, where $-2 \leq c^{\prime}-6<d^{\prime}-6$. From this, the possible values for the pair $\left(c^{\prime}-6 ; d^{\prime}-6\right)$ are $(1 ; 36)$, $(2 ; 18)$, $(3 ; 12)$, and $(4 ; 9)$, meaning in this case the pair $\left(c^{\prime} ; d^{\prime}\right)$ can be $(7 ; 42)$, $(8 ; 24)$, $(9 ; 18)$, or $(10 ; 15)$.

Thus, 6 different sets of $a^{\prime}<b^{\prime}<c^{\prime}<d^{\prime}$ satisfy the condition. Accordingly, the problem has $6 \cdot 4!=144$ different solutions, which can be obtained by all possible permutations of the quadruples $(1 ; 3 ; 4 ; 19)$, $(1 ; 3 ; 5 ; 11)$, $(1 ; 2 ; 6 ; 41)$, $(1 ; 2 ; 7 ; 23)$, $(1 ; 2 ; 8 ; 17)$, and $(1 ; 2 ; 9 ; 14)$.