$4 \cdot 7$ 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.
(26th International Mathematical Olympiad Candidate Problem, 1985)
[Solution] Since among 10 consecutive integers, there must be numbers divisible by $2^{3}, 3^{2}, 5, 7$, we have
$$
n=2^{\alpha_{1}} \cdot 3^{\alpha_{2}} \cdot 5^{\alpha_{3}} \cdot 7^{\alpha_{4}} \cdot 11^{\alpha_{5} \cdots},
$$

where $\alpha_{1} \geqslant 3, \quad \alpha_{2} \geqslant 2, \quad \alpha_{3} \geqslant 1, \quad \alpha_{4} \geqslant 1, \cdots$.
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 $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right)\left(\alpha_{4}+1\right) \geqslant 4 \cdot 3 \cdot 2 \cdot 2=48$, thus we have
$$
\left(\alpha_{5}+1\right) \cdots \leqslant 3 \text {. }
$$

Therefore, at most one $\alpha_{j}(j \geqslant 5)$ can be a positive number, and $\alpha_{j}$ can only be 1 or 2.
Considering 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 $\alpha_{1} \geqslant 3, \quad \alpha_{2} \geqslant 2, \quad \alpha_{3} \geqslant 1, \quad \alpha_{4} \geqslant 1, \quad \alpha_{5}=1$ or 2, and making $n$ as small as possible, we get
$$
\alpha_{1}=5, \quad \alpha_{2}=2, \quad \alpha_{3}=\alpha_{4}=\alpha_{5}=1 .
$$

Thus,
$$
n=2^{5} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 11=110880 \text {. }
$$

At this point, $n$ has consecutive positive divisors $1,2,3, \cdots, 11,12$.