Let $S_{m}=a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}$, and suppose $a_{i}=\frac{r_{i}}{t_{i}}, r_{i}, t_{i} \in \mathbf{Z}, t_{i}>0$, and $\left(r_{i}, t_{i}\right)=1, i=1,2, \cdots, n$.

If $a_{1}, a_{2}, \cdots, a_{n}$ are not all integers, then there must be a $t_{i}>1$. Let $p$ be a prime factor of $t_{1} t_{2} \cdots t_{n}$, and let $e_{i}$ be the largest non-negative integer such that $p^{i_{i}} \mid t_{i}$. Without loss of generality, assume
$$e_{1}=e_{2}=\cdots=e_{j}>e_{j+1} \geqslant \cdots \geqslant e_{n}$$

Let $M=\left[t_{1}, t_{2}, \cdots, t_{n}\right]$, then we can set
$$M=p^{e_{1}} \cdot N$$

where $N \in \mathbf{N}^{*}$, and $p \nmid N$. Now, since $S_{m} \in \mathbf{Z}$, it follows that
$$T_{m}=\left(a_{1} p^{e_{1}} N\right)^{m}+\left(a_{2} p^{e_{1}} N\right)^{m}+\cdots+\left(a_{n} p^{e_{1}} N\right)^{m}$$

is a multiple of $p^{m w_{1}}$.
Let $m=p^{j}-p^{j-1}\left(=\varphi\left(p^{j}\right)\right)$, and consider the terms in $T_{m}$.
For $1 \leqslant r \leqslant j$, by Euler's theorem (note that $p \nmid a_{r} p^{e_{1}} N$),
$$\left(a_{r} p^{e_{1}} N\right)^{m} \equiv\left(a_{r} p^{e_{1}} N\right)^{q\left(p^{j}\right)} \equiv 1\left(\bmod p^{j}\right)$$

While for $r>j$, $p \mid a_{r} p^{e_{1}} N$. Given that for $p \geqslant 2, j \geqslant 1$,
$$m=p^{j}-p^{j-1} \geqslant j$$

it follows that
$$\left(a_{r} p^{e_{1}} N\right)^{m} \equiv 0\left(\bmod p^{j}\right)$$

The above discussion shows that
$$T_{m}=\sum_{r=1}^{n}\left(a_{r} p^{e_{1}} N\right)^{m} \equiv j\left(\bmod p^{j}\right)$$

However, $j<p^{j}$, so $p^{j} \nmid T_{m}$. Consequently, $T_{m}$ is not a multiple of $p^{m \epsilon_{1}}$. This is a contradiction.
Therefore, $a_{1}, a_{2}, \cdots, a_{n}$ are all integers.