Sum of fractions

Assume without loss of generality that $a<b<c<d$

Then $n \neq 3$ as then the maximum of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ is

$\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{57}{60}<1$

Therefore, $a=2$ .

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

If $b=3$ , then $\frac{1}{c}+\frac{1}{d}=\frac{1}{6}$

$\frac{c+d}{cd}=\frac{1}{6}$

$6c+6d=cd$

$cd-6c-6d=0$

$(c-6)(d-6)=36$

There are four ways to factorise 36 into two factors- 1 and 36, 2 and 18, 3 and 12, and 4 and 9, so there are 4 solutions here.

If $b=4$ , then similarly, we get $(c-4)(d-4)=16$

There are two ways to factorise 16- 1 and 16, and 2 and 8, so there are two solutions here.

If $b=5$ , then the maximum of $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$ is

$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}=\frac{107}{210}=\frac{1}{2}+\frac{1}{105}$

However, we cannot make this smaller by $\frac{1}{105}$ for this to get to $\frac{1}{2}$ , so there are no solutions here.

Therefore, there are 6 solutions when $a<b<c<d$ .

However, we need to take into account the total number of permutations, so there are $6×4!=144$ solutions.

$We~ know~~~(I)~1/2+1/2=1,~~~~~~~(II)~ 1/3+2/3=1~~~~~~~(III)~1/4+3/4=1.............\\ WLOG~a<b<c<d~~~all~~integers......(*)~~~~~All~our~numerators~must~be~1...........(**)~~~~~1/a+1/b+1/c+1/d=1...........(***)\\ \because ~\text{of (*), (II) a=3 and b, c, d has to be greater than 3. So maximum we get is1/2+1/4+1/5+1/6<1. No solution by~(***).}\\ In~ the~ same~ way ~only~~ a=2~~ is~ possible. \\ \text{With a=2, maximum we can get is 1/2+1/3+1/4+1/5>1.~So~we~try~out~our~options}.\\ \Large (A)~~ a=2,~ b=3:- \\ ~ 1-(1/2+1/3)=1/6.~~~ So ~1/c+1/d=1/6.\\ Let~c=7.~~~~1/6-1/7~=1/42.~~~~~So~ (2,3,7,42)~ is~ the~ solution......\color{#3D99F6}{1}\\ Let~c=8.~~~~1/6-1/8~=1/24.~~~~~So ~(2,3,8,24)~ is~ the~ solution......\color{#3D99F6}{2}\\ Let~c=9.~~~~1/6-1/9~=1/18.~~~~~So~ (2,3,9,18)~ is~ the~ solution......\color{#3D99F6}{3}\\ Let~c=10.~~~1/6-1/10=1/15.~~~So~ (2,3,10,15)~ is ~the ~solution.....\color{#3D99F6}{4}\\ Let~c=11.~~~1/6-1/11=5/66~~but~by~(**)~it~ is~ not~valid.\\ Let~c=12.~~~1/6-1/12=1/12~~but~c=d,~not~valid, ~by~(*).\\ For~c>12,~~~~1/c+1/d<1/6,~~so~no~more~solutions~with~a=2.~b=3.\\ \Large (B) ~~a=2,~ b=4:- \\ 1-(1/2+1/4)=1/4.~~~ So~ 1/c+1/d=1/4.\\ Let~c=5.~~~~1/4-1/5~=1/20.~~~~~So~ (2,4,5,20)~ is~ the~ solution......\color{#3D99F6}{5}\\ Let~c=6.~~~~1/4-1/6~=1/12.~~~~~So~ (2,4,6,12) ~is ~the ~solution......\color{#3D99F6}{6}\\ Let~c=7.~~~~1/4-1/7~=3/28~~but~by~(*)~it~ is ~not~valid.\\ Let~c=8.~~~~1/4-1/8~=1/8~~but~c=d,~not~valid,~by~(*)\\ For~c>8,~~~~1/c+1/d<1/4,~~so~no~more~solutions~with~a=2.~b=4.\\ \Large (C)~~ a=2,~~ b=5:- \\ 1-(1/2+1/5)=3/10.~~~ So ~1/c+1/d=3/10.\\ Let~c=6.~~~~3/10-1/6~=2/15~~but~by~(*) ~it~is~ not~valid.\\ For~c>6,~~~~1/c+1/d<3/10,~~so~no~more~solutions~with~a=2.~b=5.\\ \implies~(C)~has~no~ solution.\\ \Large (D)~~ a=2,~ b=6:- \\ a=2, b=c=d,~~\implies~1/2+1/6+1/6+1/6=1.\\ But~our~c,~and~d~are~going~to~be~GREATER~than~6~and~so~for~b\geq 6~no~solution.\\ So~~there~are~6~~combinations.~But~permutations~would~be~~=~~6*4!=\Large \color{#D61F06}{144}.$