6.2. How many natural numbers $\mathrm{N}$ greater than 300 exist such that among the numbers $4 \mathrm{~N}, \mathrm{~N}-$ $300, N+45,2 N$ exactly two are four-digit numbers?
Answer: 5410

Solution. Note that $2 \mathrm{~N}>\mathrm{N}+15$, and if we write the numbers in ascending order, we get $\mathrm{N}-300, \mathrm{~N}+45,2 \mathrm{~N}, 4 \mathrm{~N}$. Only two consecutive numbers can be four-digit. If the four-digit numbers are $\mathrm{N}-300$ and $\mathrm{N}+15$, then $2 \mathrm{~N}$ must be a five-digit number, i.e., $\mathrm{N} \geq 5000$. At the same time, N is less than 9955 (so that N + 45 is a four-digit number), i.e., N is from 5000 to 9954, 4955 options. If $\mathrm{N}+45,2 \mathrm{~N}$ are four-digit, then $\mathrm{N}-300$ is a three-digit number, $\mathrm{N} \leq 999+$ $300=1399$, but then $4 \mathrm{~N}$ cannot be a five-digit number. If the numbers $2 \mathrm{~N}$ and $4 \mathrm{~N}$ are four-digit, then $\mathrm{N} \geq 500$. At the same time, $\mathrm{N}+45$ must be a three-digit number, from which $\mathrm{N} \leq 954$, that is, $\mathrm{N}$ takes values from 500 to 954, another 455 options. In total $4955+455=5410$ options.