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: $${1 \over a} > {1
\over b} > {1\over c}$$Consequently, the original equality: $${1
\over a} + {1 \over b} + {1\over c} = 1$$transforms, keeping into
account (1) (${1\over a}> {1\over b}$ and ${1\over a}>
{1\over c}$) into: $${3\over a} > 1$$ i.e. $a< 3$.

Because $a$ is integer, and because it cannot be 1 (if it
would, then: ${1\over b} + {1\over c}=0$) it means that the only
one possibility is $a = 2$. Then: $${1\over 2}+{1\over b}+{1\over
c} = 1 \mbox{ or } {1\over b}+{1\over c} = {1\over 2}$$Because from
(1): ${1\over b} > {1\over c}$, (3) transforms into: ${2\over
b}> {1\over 2}$ 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:
${1\over a} + {1\over b} +{1\over c} + {1\over d } = 1$

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

#### ${1\over 2}+{1\over 3}+{1\over
4}+{1\over d} = 1$ has no solution for $d$ integer.

#### c = 5

#### ${1\over 2}+{1\over 3}+{1\over
5}+{1\over d} = 1$ has no solution for $d$ integer.

#### c = 6

#### ${1\over 2}+{1\over 3}+{1\over
6}+{1\over d} = 1$ has no solution for $d$ integer.

#### c = 7

#### ${1\over 2}+{1\over 3}+{1\over
7}+{1\over d} = 1$ gives $d=42$.

#### c = 8

#### ${1\over 2}+{1\over 3}+{1\over
8}+{1\over d} = 1$ gives $d=24$.

#### c = 9

#### ${1\over 2}+{1\over 3}+{1\over
9}+{1\over d} = 1$ gives $d=18$.

#### c = 10

#### ${1\over 2}+{1\over 3}+{1\over
10}+{1\over d} = 1$ gives $d=15$.

#### c = 11

#### ${1\over 2}+{1\over 3}+{1\over
11}+{1\over d} = 1$ 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

#### ${1\over 2}+{1\over 4}+{1\over
5}+{1\over d} = 1$ gives $d=20$.

#### c = 6

#### ${1\over 2}+{1\over 4}+{1\over
6}+{1\over d} = 1$ gives $d=12$.

#### c = 7

#### ${1\over 2}+{1\over 4}+{1\over
7}+{1\over d} = 1$ hasn'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 |