Here is Andrei Lazanu's solution. Andrei is 12 years old and a pupil at School Number 205, Bucharest, Romania. You don't need much mathematical knowledge to solve this problem but you do need a lot of mathematical thinking. Well done Andrei.

I started from the relation 0 < a < b < c, so a, b and c are positive. This means that:
 1
a
 >  1
b
 >  1
c
(1)

Consequently, the original equality:
 1
a
 +  1
b
 +  1
c
 = 1
transforms, keeping into account (1) (1/a > 1/b and 1/a > 1/c) into:
 3
a
 > 1
(2)

i.e. a < 3.

Because a is integer, and because it cannot be 1 (if it would, then: 1/b + 1/c=0) it means that the only one possibility is a = 2.

Then:
 1
2
 +  1
b
 +  1
c
 = 1 or  1
b
 +  1
c
 =  1
2
(3)

Because from (1): 1/b > 1/c, (3) transforms into: 2/b > 1/2 or b < 4. Because b > a and a = 2, there is only one possibility for b, b = 3. Substituting into the original relation, I obtained c = 6.

With four numbers, I followed the same procedure. From:
 1
a
 +  1
b
 +  1
c
 +  1
d
= 1

I obtained: 4/a > 1, so a < 4. Again a cannot be 1, so it must be either 2 or 3. I analyse the possibilities one by one.

a = 2

Using again the relation of order between b, c and d I found b < 6. Because b > a, I have to analyse b = 3, b = 4 and b = 5.

b = 3

From the relation between c and d, and substituting the values for a and b I obtained c < 12, it means it could be 4, 5, 6, 7, 8, 9, 10 or 11.

c = 4


 1
2
 +  1
3
 +  1
4
 +  1
d
 = 1 has no solution for d integer.

c = 5


 1
2
 +  1
3
 +  1
5
 +  1
d
 = 1 has no solution for d integer.

c = 6


 1
2
 +  1
3
 +  1
6
 +  1
d
 = 1 has no solution for d integer.

c = 7


 1
2
 +  1
3
 +  1
7
 +  1
d
 = 1 gives d=42.

c = 8


 1
2
 +  1
3
 +  1
8
 +  1
d
 = 1 gives d=24.

c = 9


 1
2
 +  1
3
 +  1
9
 +  1
d
 = 1 gives d=18.

c = 10


 1
2
 +  1
3
 +  1
10
 +  1
d
 = 1 gives d=15.

c = 11


 1
2
 +  1
3
 +  1
11
 +  1
d
 = 1 has no solution for d integer.

b = 4

Repeating the same procedure, I obtained: c < 8, and because c > b, c could be 5, 6 and 7.

c = 5


 1
2
 +  1
4
 +  1
5
 +  1
d
 = 1 gives d=20.

c = 6


 1
2
 +  1
4
 +  1
6
 +  1
d
 = 1 gives d=12.

c = 7


 1
2
 +  1
4
 +  1
7
 +  1
d
 = 1 hasn < quote/ > t a solution for d integer.

b = 5

I obtained: 3c < 20, and consequently only c = 6 is possible. This hasn't a solution for d.

a = 3

For b, I obtained: 2b < 9, and because b > a, it is possible only b = 4. Then, I obtained 5c < 24, i. e. without solution because c > b.

Finally, the good solutions are:

a 2 2 2 2 2 2
b 3 3 3 3 4 4
c 7 8 9 10 5 6
d 42 24 18 15 20 12