The "grand" idea of this question is to use Bèzout theorem. So note that $\sum_{i=1}^{k^m}\frac{1}{(ak)^m}=\frac{1}{a^m}$, for any positive interger $k$ . Then, in fact, if a=1, we can write 1 as the sum of $k^m$ numbers of the form $\frac{1}{k^m}$: $1=\sum_{i=1}^{k^m}\frac{1}{(k)^m}$, with $n_0=k^m$. Now, using the last idea, we can write $\frac{1}{k^m}$ as $\sum_{i=1}^{p^m}\frac{1}{(pk)^m}$(1), for some arbitrary positive interger $p$ adding $p^m -1$ terms to this sum or write $\frac{1}{k^m}$ as $\sum_{i=1}^{q^m}\frac{1}{(qk)^m} $(2), for some arbitrary interger $q$ adding $q^m -1$ terms to the sum. Then, the resulting $n$ is equals to: $k^m +(p^m-1)u +(q^m-1)v$, after replacing (1) $u$ times and (2) $v$ times. We only need to have $gdc(p^m-1,q^m-1)=1$, because for $gcd(A,B)=1$ and $a,b$ varying on non-negative intergers, then $n= aA+bB$ always have solution if $n>AB-A-B$ [Chicken McNugget Theorem]. Then we just need to have $gcd(p^m - 1,q^m-1)=1$. Take $q=l(p^m -1)$ for some positive interger $l$ and it is over, as desired. It is a very technical question for a P4.

"Now, using the last idea, we can write $\frac{1}{k^m}$ as $\sum_{i=1}^{p^m}\frac{1}{(pk)^m}$ (1) adding $p^m -1$ terms to this sum or write $\frac{1}{k^m}$ as $\sum_{i=1}^{q^m}\frac{1}{(qk)^m} $(2), adding $q^m -1$ terms to the sum. Then, the resulting $n$ is equals to: $k^m +(p^m-1)u +(q^m-1)v$, after replacing (1) $u$ times and (2) $v$ times." What's $n$? Why $n=k^m +(p^m-1)u +(q^m-1)v$? Can you explain this text?

analysis90 wrote: "Now, using the last idea, we can write $\frac{1}{k^m}$ as $\sum_{i=1}^{p^m}\frac{1}{(pk)^m}$ (1) adding $p^m -1$ terms to this sum or write $\frac{1}{k^m}$ as $\sum_{i=1}^{q^m}\frac{1}{(qk)^m} $(2), adding $q^m -1$ terms to the sum. Then, the resulting $n$ is equals to: $k^m +(p^m-1)u +(q^m-1)v$, after replacing (1) $u$ times and (2) $v$ times." What's $n$? Why $n=k^m +(p^m-1)u +(q^m-1)v$? Can you explain this text? Of course! It's probably not 100% clear. However, note that $\frac{1}{k^m}$ can be expanded to $\sum_{i=1}^{p^m}\frac{1}{(pk)^m}$ for any pair of arbitrary positive intergers $(p, k)$. It means that for a fixed $k$, we can choose $p$ in order to increase the number of terms of this sum: $\sum_{i=1}^{k^m}\frac{1}{(k)^m}$. And it is clear that, as $\frac{1}{k^m}$ has only one term and $\sum_{i=1}^{p^m}\frac{1}{(pk)^m}$ has $p^m $ terms, we added $p^m -1$ terms to the sum. It is the same for $q$. Then, take $n_0$ as the number of terms of $\sum_{i=1}^{k^m}\frac{1}{(k)^m}=1$, obviously $n_0=k^m$. Now take $n$ as the number of terms of the last sum after doing the operation (1) $u$ times and (2) $v$ times as I described in my solution, hence $n=n_0 +(p^m-1)u +(q^m-1)v=k^m +(p^m-1)u +(q^m-1)v$, since $n_0=k^m$ and that these operations add $p^m -1$ or $q^m - 1$ terms to the sum. Ps: $u$ and $v$ can be as big as necessary, because the operations only require a term, it does not depends on which term is.

Surprisingly, it is an old problem that has appeared in 250 Problems in Elementary Number Theory - Sierpinski (1970)

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