Let $d_{1}=1, d_{2}=n-1, d_{3}=d_{4}=\cdots=d_{n}=n$, then $d_{1}+d_{2}+\cdots+d_{n}=1+$ $(n-1)+(n-2) \cdot n=(n-1) n$, which is divisible by every $d_{i}(i=1,2, \cdots, n)$, and the greatest common divisor of $d_{1}, d_{2}, \cdots, d_{n}$ is 1, at this time $d_{1} d_{2} \cdots d_{n}=(n-1) n^{n-2}$. If $d_{1} d_{2} \cdots d_{n}$ divides $\left(d_{1}+\cdots+d_{n}\right)^{k}$, i.e., $(n-1) n^{n-2} \mid[(n-1) n]^{k}$, by $(n, n-1)=1$ we get $k \geqslant n-2$.

Conversely, we prove: If the greatest common divisor of $d_{1}, d_{2}, \cdots, d_{n}$ is 1, and $d_{i}$ divides $\left(d_{1}+d_{2}+\cdots+d_{n}\right)$ $(i=1,2, \cdots, n)$, then $d_{1} d_{2} \cdots d_{n} \mid\left(d_{1}+d_{2}+\cdots+d_{n}\right)^{n-2}$.

In fact, let $p$ be any prime factor of $d_{1} d_{2} \cdots d_{n}$, then there exists a positive integer $k$, such that at least one of $d_{1}, d_{2}, \cdots, d_{n}$ is divisible by $p^{k}$, and no $d_{1}, d_{2}, \cdots, d_{n}$ is divisible by $p^{k+1}$. Thus, by $d_{i} \mid\left(d_{1}+d_{2}+\cdots+d_{n}\right)$ we get $p^{k} \mid\left(d_{1}+d_{2}+\cdots+d_{n}\right)$, hence $p^{k(n-2)} \mid\left(d_{1}+d_{2}+\cdots+d_{n}\right)^{n-2}$. Since the greatest common divisor of $d_{1}, d_{2}, \cdots, d_{n}$ is 1, there must be some $d_{j}$ not divisible by $p$. Also, since $p$ divides $d_{1}+$ $d_{2}+\cdots+d_{n}$, at least two of $d_{1}, d_{2}, \cdots, d_{n}$ are not divisible by $p$, so the highest power of $p$ dividing $d_{1} d_{2} \cdots d_{n}$ is at most $p^{k(n-2)}$. Repeating the above discussion for each prime factor of $d_{1}, d_{2}, \cdots, d_{n}$, we know that $d_{1} d_{2} \cdots$ $d_{n}$ divides $\left(d_{1}+d_{2}+\cdots+d_{n}\right)^{n-2}$.
In summary, the smallest positive integer $k=n-2$.
Example 3 Given a positive integer $n \geqslant 3$, let $d_{1}, d_{2}, \cdots, d_{n}$ be positive integers, whose greatest common divisor is 1, and $d_{i}$ divides $d_{1}+d_{2}+\cdots+d_{n}(i=1,2, \cdots, n)$. Find the smallest positive integer $k$, such that $d_{1} d_{2} \cdots d_{n}$ divides $\left(d_{1}+\right.$ $\left.d_{2}+\cdots+d_{n}\right)^{k}$.