4.3. [6] Find all finite arithmetic progressions with sum 1, each term of which is of the form $\frac{1}{k}$, where $k$ is a natural number.
Solution. By multiplying each term of the sequence by the LCM of the denominators, we will obtain an increasing arithmetic progression $\left\{a_{i}\right\}$ of $n$ natural numbers, whose sum $S$ is divisible by each term of the sequence. The greatest common divisor of the terms of this sequence is 1, and therefore the difference $d$ is coprime with each term of the sequence.

We will consider two cases.

1) $n=2 m$. Then $S=\frac{n\left(a_{m}+a_{m+1}\right)}{2}=n a_{m+1}-m d$, from which $m d$, and therefore $m$, is a multiple of $a_{m+1}$. But this is impossible, since $a_{m+1}>m$.
2) $n=2 m+1$. Then $S=n a_{m+1}=n a_{m}+n d$, from which $n=2 m+1$ is a multiple of $a_{m}$. Since $a_{m} \geq m$, this is possible only if $a_{m}=m$ and $m=1$. From this, we obtain the only solution: $\frac{1}{6}, \frac{1}{3}, \frac{1}{2}$.