Example 8 (IMO-26 Shortlist) Find the smallest positive integer $n$ that satisfies:
(1) $n$ has exactly 144 different positive divisors;
(2) Among the positive divisors of $n$, there are 10 consecutive integers.
Since among 10 consecutive integers, there must be numbers divisible by $2^{3}, 3^{2}, 5, 7$, it follows that $\alpha_{1} \geqslant 3, \alpha_{2} \geqslant 2, \alpha_{3} \geqslant 1, \alpha_{4} \geqslant 1, \cdots$.
Since the number of positive divisors of $n$ is
$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right)\left(\alpha_{5}+1\right) \cdots=144,
$$

and $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right) \geqslant 4 \cdot 3 \cdot 2 \cdot 2=48$,
it follows that $\left(\alpha_{5}+1\right) \cdots \leqslant 3$.
Therefore, at most one $\alpha_{j}(j \geqslant 5)$ can be a positive number, and $\alpha_{j}$ can only be 1 or 2.
Consider the equation $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right)\left(\alpha_{5}+1\right)=144$ for all solutions with $\alpha_{1} \geqslant 3, \alpha_{2} \geqslant 2, \alpha_{3} \geqslant 1, \alpha_{4} \geqslant 1, \alpha_{5}=1$ or 2, and make $n$ as small as possible. This leads to
$$
\alpha_{1}=5, \alpha_{2}=2, \alpha_{3}=\alpha_{4}=\alpha_{5}=1 .
$$

Thus, $n=2^{5} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 11=110880$.
In this case, $n$ has consecutive positive divisors $1,2,3, \cdots, 11,12$.