Let's determine the smallest positive integer that has 144 positive divisors, and among these, there are ten consecutive ones.
Among ten consecutive integers, there is always one divisible by 5, one by 7, one by 8, and one by 9. The number we are looking for is therefore also divisible by these numbers. Since these four numbers are pairwise relatively prime, the number we are looking for is also divisible by their product, $2^{3} \cdot 3^{2} \cdot 5 \cdot 7=2520$. Conversely: If an integer is divisible by 2520, then it is clearly divisible by all divisors of 2520, and among the latter, there exist ten consecutive numbers, namely $1,2,3, \ldots, 10$. Our task is therefore to find the smallest positive integer $n$ that is divisible by 2520 and has 144 positive divisors.

Let the prime factorization of $n$ be $n=p_{1}^{a_{1}} \cdot p_{2}^{a_{2}} \cdot \ldots \cdot p_{k}^{a_{k}}$, where $p_{1}<p_{2}<\ldots<p_{k}$. It is known that the number of (positive) divisors of $n$ is $\left(a_{1}+1\right)\left(a_{2}+1\right) \cdot \ldots \cdot\left(a_{k}+1\right)$. We will show that $a_{1} \geqq a_{2} \geqq \ldots \geqq a_{k}$. Suppose, for example, that $a_{1}<a_{2}$; then the number $n^{\prime}=p_{1}^{a_{2}} \cdot p_{2}^{a_{1}} \cdot p_{3}^{a_{3}} \cdot \ldots \cdot p_{k}^{a_{k}}$ has the same number of divisors as $n$, and $n^{\prime}$ is also divisible by 2520, since in its prime factorization, the relationship between the exponents is the opposite of the relationship between the corresponding primes. However, since $n^{\prime}<n$, we arrive at a contradiction with the assumption that $n$ is the smallest number with the desired property. From the minimality of $n$, it also follows that in the prime factorization of $n$, $p_{1}=2, p_{2}=3$, $p_{3}=5, p_{4}=7$, and generally $p_{i}$ is the $i$-th prime. Since $n$ is divisible by 2520, it follows that $a_{1} \geqq 3, a_{2} \geqq 2, a_{3} \geqq 1$ and $a_{4} \geqq 1$. Based on the relationship $\left(a_{1}+1\right) \cdot \ldots \cdot\left(a_{k}+1\right)=144$, we therefore need to express 144 as a product of at least four factors (greater than 1) such that one of the factors is at least 4 and another is at least 3. This can be done in eight ways (see the table below).

| $\left(a_{1}+1\right) \ldots\left(a_{k}+1\right)$ | $n$ |
| :--- | :--- |
| $4 \cdot 4 \cdot 3 \cdot 3$ | $2^{3} \cdot 3^{3} \cdot 5^{2} \cdot 7^{2}=264600$ |
| $4 \cdot 3 \cdot 3 \cdot 2 \cdot 2$ | $2^{3} \cdot 3^{2} \cdot 5^{2} \cdot 7 \cdot 11=138600$ |
| $6 \cdot 4 \cdot 3 \cdot 2$ | $2^{5} \cdot 3^{3} \cdot 5^{2} \cdot 7=151200$ |
| $6 \cdot 6 \cdot 2 \cdot 2$ | $2^{5} \cdot 3^{3} \cdot 5 \cdot 7=272160$ |
| $6 \cdot 3 \cdot 2 \cdot 2 \cdot 2$ | $2^{5} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 11=110880$ |
| $8 \cdot 3 \cdot 3 \cdot 2$ | $2^{7} \cdot 3^{2} \cdot 5^{2} \cdot 7=201600$ |
| $9 \cdot 4 \cdot 2 \cdot 2$ | $2^{8} \cdot 3^{3} \cdot 5 \cdot 7=241920$ |
| $12 \cdot 3 \cdot 2 \cdot 2$ | $2^{11} \cdot 3^{2} \cdot 5 \cdot 7=645120$ |

Among these numbers, the smallest one we are looking for is 110880.