6.3. How many natural numbers $\mathrm{N}$ greater than 700 exist such that among the numbers $3 \mathrm{~N}, \mathrm{~N}-$ $700, N+35,2 N$ exactly two are four-digit numbers?
Answer: 5229

Solution. Note that $2 \mathrm{~N}>\mathrm{N}+35$, and if we write the numbers in ascending order, we get $\mathrm{N}-700, \mathrm{~N}+35,2 \mathrm{~N}, 3 \mathrm{~N}$. Only two consecutive numbers can be four-digit. If the four-digit numbers are $\mathrm{N}-700$ and $\mathrm{N}+35$, then $2 \mathrm{~N}$ must be a five-digit number, i.e., $\mathrm{N} \geq 5000$. At the same time, N is less than 9965 (so that N + 35 is a four-digit number), i.e., N is from 5000 to 9964, 4965 options. If $\mathrm{N}+35,2 \mathrm{~N}$ are four-digit, then $\mathrm{N}-700$ is a three-digit number, $\mathrm{N} \leq 999+$ $700=1699$, but then $3 \mathrm{~N}$ cannot be a five-digit number. If the numbers $2 \mathrm{~N}$ and $3 \mathrm{~N}$ are four-digit, then $\mathrm{N} \geq 500$, but according to the condition $\mathrm{N}>700$. At the same time, $\mathrm{N}+35$ must be a three-digit number, from which $N \leq 964$, so $N$ takes values from 701 to 964, another 264 options. In total $4965+264=5229$ options.