4. Among the integers from 1 to 2020, there are 1616 numbers that are not multiples of 5. Divide these 1616 numbers into several groups (the numbers in each group do not have to be the same), such that the difference (larger number minus smaller number) between any two numbers in each group is a prime number. What is the minimum number of groups needed?
【Answer】404
【Analysis】Idea: Fewer groups, with as many numbers as possible in each group. We start enumerating from the minimum number that satisfies the condition.
2 numbers per group is obviously possible, just separate odd and even numbers; for 3 numbers in a group: 1, 3, 6 (essentially three numbers with differences of 2 and 3); for 4 numbers in a group: 1, 3, 6, 8; but for 5 numbers, after 1, 3, 6, 8, it is impossible to form a valid group. Upon careful consideration, the first 4 numbers consist of 2 odd and 2 even numbers. No matter what the 5th number is, it will form 2 different even numbers, which cannot all be prime numbers. Therefore, the maximum number of numbers in a group is 4. Next, we will argue this.

Argument: For any 5 natural numbers $A, B, C, D, E$, let $X_{1}=B-A, X_{2}=C-B, X_{3}=D-C, X_{4}=E-D$. In the sequence $X_{1} \sim X_{2} \sim X_{3} \sim X_{4}$, adjacent numbers must have different parity, otherwise the sum of adjacent numbers would be an even number greater than 2, which is not a prime number. If $X_{1}$ is odd, then $X_{1}+X_{2}+X_{3}$ is an even number greater than 2; if $X_{1}$ is even, then $X_{2}+X_{3}+X_{4}$ is an even number greater than 2. Therefore, the number of numbers in a group does not exceed 4, so the number of groups does not exceed 404 groups.

Construction: Consider the groups $\{1,3,6,8\},\{2,4,7,9\},\{11,13,16,18\}$ and $\{12,14,17,19\}, \ldots, \{2011,2013,2016,2018\}$ and $\{2012,2014, 2017, 2019\}$. Each group contains 4 numbers, totaling $1616 \div 4=404$ groups, which is a valid grouping.