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:

${\displaystyle \frac{1}{a}>\frac{1}{b}>\frac{1}{c}}$

${\displaystyle (1)}$

Consequently, the original equality:

${\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1}$

transforms, keeping into account (1) ( $\frac{1}{a}>\frac{1}{b}$ and $\frac{1}{a}>\frac{1}{c}$) into:

${\displaystyle \frac{3}{a}>1}$

${\displaystyle (2)}$

i.e. $a<3$.

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

Then:

${\displaystyle \frac{1}{2}+\frac{1}{b}+\frac{1}{c}=1\text{or}\frac{1}{b}+\frac{1}{c}=\frac{1}{2}}$

${\displaystyle (3)}$

Because from (1): $\frac{1}{b}>\frac{1}{c}$, (3) transforms into: $\frac{2}{b}>\frac{1}{2}$ 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:

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

I obtained: $\frac{4}{a}>1$, 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

c = 4

c = 5

c = 6

c = 7

c = 8

c = 9

c = 10

c = 11

b = 4

c = 5

c = 6

c = 7

b = 5

a = 3

Finally, the good solutions are: