WakeUp wrote: Determine all positive integers in the form $a<b<c<d$ with the property that each of them divides the sum of the other three. So $(a+b+c)=kd$ but $a+b+c<3d$ and so $k\in\{1,2\}$ 1) $k=1$ and $d=a+b+c$ Then $c|a+b+(a+b+c)$ and so $c|2(a+b)$ and $2(a+b)=k_1c$ but $2(a+b)<4c$ and so $k_1\in\{1,2,3\}$ 1.1) $k_1=1$ and $c=2(a+b)$ The numbers are $a,b,2a+2b,3a+3b$ And $b|6a+5b$ and so $b|6a$ and, $a|6b$ Let then $a=a_1u$ and $b=b_1u$ with $a_1<b_1$ coprime. We get $b_1|6$ and $a_1|6$ and so $(a_1,b_1)$ is $(1,2),(1,3),(1,6),(2,3),(2,6),(3,6)$ And solutions : $(a,b,c,d)=(u,2u,6u,9u)$ $(a,b,c,d)=(u,3u,8u,12u)$ $(a,b,c,d)=(u,6u,14u,21u)$ $(a,b,c,d)=(2u,3u,10u,15u)$ 1.2) $k_1=2$ and $c=a+b$ The numbers are $a,b,a+b,2a+2b$ And $b|4a+3b$ and so $b|4a$ and $a|4b$ Let then $a=a_1u$ and $b=b_1u$ with $a_1<b_1$ coprime. We get $b_1|4$ and $a_1|4$ and so $(a_1,b_1)$ is $(1,2),(1,4)$ And solutions : $(a,b,c,d)=(u,2u,3u,6u)$ $(a,b,c,d)=(u,4u,5u,10u)$ 1.3) $k_1=3$ and $c=\frac{2a+2b}3$ The numbers are $a,3t-a,2t,5t$ with $2t<2a<3t$ And $a|10t-a$ and so $a|10t$ and $10t=ak'$ and since $\frac{20}3a<10t<10a$, we get $10t=7a$ or $10t=8a$ or $10t=9a$ $10t=7a$ implies $t=7p$ and $a=10p$ and the numbers are $(10p,11p,14p,35p)$ which is not a solution $10t=8a$ implies $t=4p$ and $a=5p$ and the numbers are $(5p,7p,8p,20p)$ which is not a solution $10t=9a$ implies $t=9p$ and $a=10p$ and the numbers are $(10p,17p,18p,45p)$ which is not a solution 2) $k=2$ and $d=\frac{a+b+c}2$ Then $c|a+b+\frac{a+b+c}2$ and so $\frac 32(a+b)+\frac c2=k_1c$ and so $\frac 32(a+b)=(k_1-\frac 12)c$ And since $a+b<2c$, we get $k_1<3+\frac 12$ and so $k_1\in\{1,2,3\}$ 2.1) $k_1=1$ and $c=3(a+b)$ Then $d=2(a+b)<c$ and so no solution 2.2) $k_1=2$ and $c=a+b$ Then $d=a+b=c$ and so no solution 2.3) $k_1=3$ and $5c=3(a+b)$ The numbers are $a,5t-a,3t,4t$ with $4t<2a<5t$ So $a|12t-a$ and so $a|12t$ and so $12t=k_2a$ but $\frac{24}5a<12t<6a$ and so $12t=5a$ The numbers are $(12p,13p,15p,20p)$ which is not a solution. Hence the set of basis solutions (all other are obtained thru multiplication by any positive integer) : $(a,b,c,d)=(1,2,3,6)$ $(a,b,c,d)=(1,2,6,9)$ $(a,b,c,d)=(1,3,8,12)$ $(a,b,c,d)=(1,4,5,10)$ $(a,b,c,d)=(1,6,14,21)$ $(a,b,c,d)=(2,3,10,15)$

I don’t know if there is any better solution but apparently they just wanted us to casework 6 times.