Example 7 Find the smallest positive integer $n$, which has exactly 144 different positive divisors, and among them, there are 10 consecutive divisors.
Solution: Since the positive divisors of $n$ include 10 consecutive integers, it must have multiples of $2^{3}, 3^{2}, 5, 7$, therefore, $n$ must be a multiple of $2^{3} \cdot 3^{2} \cdot 5 \cdot 7$.
Let the standard factorization of $n$ be $n=2^{r_{1}} \cdot 3^{r_{2}} \cdot 5^{r_{3}} \cdot 7^{r_{4}} \cdot \cdots \cdot p_{k}^{r_{k}}$, then
$$
r_{1} \geqslant 3, r_{2} \geqslant 2, r_{3} \geqslant 1, r_{4} \geqslant 1 \text {. }
$$

Also, the number of positive divisors of $n$, $d(n)=\left(r_{1}+1\right)\left(r_{2}+1\right) \cdots\left(r_{k}+1\right)=144$. And $\left(r_{1}+1\right)\left(r_{2}+1\right)\left(r_{3}+1\right)\left(r_{4}+1\right) \geqslant 4 \times 3 \times 2 \times 2=48$, therefore $\left(r_{5}+1\right)\left(r_{6}+1\right) \cdots\left(r_{k}+1\right) \leqslant 3$.

Thus, among $r_{5}, r_{6}, \cdots, r_{k}$, there is at most one that is not zero. To make $n$ the smallest, then $k=5,0 \leqslant a_{5} \leqslant 2$, so the form of $n$ is
$$
n=2^{r_{1}} \cdot 3^{r_{2}} \cdot 5^{r_{3}} \cdot 7^{r_{4}} \cdot 11^{r_{5}}
$$

Here $r_{1} \geqslant 3, r_{2} \geqslant 2, r_{3} \geqslant 1, r_{4} \geqslant 1,0 \leqslant r_{5} \leqslant 2$.
Thus, $\left(r_{1}+1\right)\left(r_{2}+1\right)\left(r_{3}+1\right)\left(r_{4}+1\right)=144$
or $\left(r_{1}+1\right)\left(r_{2}+1\right)\left(r_{3}+1\right)\left(r_{4}+1\right)\left(r_{5}+1\right)=144$
Obviously, when $r_{1} \geqslant r_{2} \geqslant r_{3} \geqslant r_{4} \geqslant r_{5}$, $n$ is the smallest.
By $144=2 \times 2 \times 2 \times 2 \times 3 \times 3$, trying to find the array $\left(r_{1}, r_{2}, r_{2}\right.$, $\left.r_{3}, r_{4}, r_{5}\right)$ that satisfies the above, we get the array $(5,2,1,1,1)$ which can make $n$ the smallest. Thus, the smallest $n=2^{5} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 11=110880$.