6.1. How many natural numbers $\mathrm{N}$ greater than 900 exist such that among the numbers $3 \mathrm{~N}, \mathrm{~N}-$ $900, N+15,2 N$ exactly two are four-digit numbers?
Answer: 5069

Solution. Note that $2 \mathrm{~N}>\mathrm{N}+15$, and if we write the numbers in ascending order, we get $\mathrm{N}-900, \mathrm{~N}+15,2 \mathrm{~N}, 3 \mathrm{~N}$. Four-digit numbers can only be two consecutive ones. If the four-digit numbers are $\mathrm{N}-900$ and $\mathrm{N}+15$, then $2 \mathrm{~N}$ must be a five-digit number, i.e., $\mathrm{N} \geq 5000$. At the same time, $\mathrm{N}$ is less than 9985 (so that $\mathrm{N}+15$ is a four-digit number), i.e., $\mathrm{N}$ is from 5000 to 9984, 4985 options. If $\mathrm{N}+15,2 \mathrm{~N}$ are four-digit numbers, then $\mathrm{N}-900$ is a three-digit number, $\mathrm{N} \leq 999+$ $900=1899$, but then $3 \mathrm{~N}$ cannot be a five-digit number. If the numbers $2 \mathrm{~N}$ and $3 \mathrm{~N}$ are four-digit numbers, then $\mathrm{N} \geq 500$, but according to the condition $\mathrm{N}>900$. At the same time, $\mathrm{N}+15$ must be a three-digit number, hence $N \leq 984$, so $N$ takes values from 901 to 984, another 84 options. In total $4985+84=5069$ options.