Provide four consecutive natural numbers, each of which has a square divisor greater than 1.
The "most stubborn" method is to take the multiples of the squares of prime numbers in sequence, and look for four consecutive numbers among them. Fortunately, the smallest such solution is quite small:

$$
242=11^{2} \cdot 2, \quad 243=3^{2} \cdot 27, \quad 244=2^{2} \cdot 61, \quad 245=7^{2} \cdot 5
$$

thus the process ends relatively quickly.

A method that requires fewer trials is as follows. Numbers ending in 24 are divisible by $2^{2}$, and those ending in 25 are divisible by $5^{2}$. Therefore, if we look for the sequence of four numbers in the form

$$
900 k+24, \quad 900 k+25, \quad 900 k+26, \quad 900 k+27
$$

then the first term of the sequence is divisible by 4, the second term by 25, and the last term by 9. We would like to choose the value of $k$ such that the third term, $900 k+26$, is divisible by 49. However, $900 k+26=18 k \cdot 49+18 k+26$ is exactly divisible by 49 if $49 \mid 18 k+26$, which is true, for example, when $k=4$. From this, we get the sequence of four numbers

$$
3624=2^{2} \cdot 906, \quad 3625=5^{2} \cdot 145, \quad 3626=7^{2} \cdot 74, \quad 3627=3^{2} \cdot 403
$$

Remark. It is also true that for any $n$, there are $n$ consecutive integers, each of which has a divisor that is a square of a number greater than 1. This is a consequence of the following, so-called Chinese Remainder Theorem: let $a_{1}, a_{2}, \ldots, a_{n}$ be pairwise coprime, and $b_{1}, b_{2}, \ldots, b_{n}$ be any integers. Then there is an integer $x$ such that for every $k$ between 1 and $n$, $x+b_{k}$ is divisible by $a_{k}$. The theorem gives our statement with the following assignment: let $a_{k}$ be the square of the $k$-th prime number, and $b_{k}=k$. Then $x+1, x+2, \ldots, x+n$ are the desired sequence of $n$ numbers.