Solution:

We can narrow down her possible choices quickly:
- If Bob chose "one," Aerith would have to have to say "one times one", so her two letters must both be in the word "one".
- If Bob chose "three," she would have to say "one times three", so one of her letters must be in "three" and thus must be "e".
- If Bob chose "four," she could not have said "two times two" as neither word contains "e", so she must have said "one times four". Thus her second letter must be "o".
Thus, Aerith has letters "o" and "e".

We now show that all positive integers can be represented this way. We note that any number containing either one of "o" or "e" works, because multiplying itself by "one" will supply the other letter. As such,
- any number 1000 or higher works because it must contain "o" ("thousand", "million", "billion", ...), and
- any number from 100 to 999 works, because it must contain "hundred",
- 10 through 19 work, because all of "ten", "eleven", "twelve", and "thirteen" through "nineteen" contain "e",
- any other number ending in digits other than 0 or 6 works, because all of "one", "two" "three", "four" "five", "seven", "eight" and "nine" also contain either "e" or "o",
- all numbers ending in 6 can be written as "two" times a number ending in "three", "eight", "thirteen" or "eighteen", all of which contain "e",
- 30, 50, and 60 can be factored as "two" times "fifteen", "twentyfive", or "twentyfive", respectively, and
- the remaining multiples of 10 , i.e., "twenty", "forty", "seventy", "eighty" and "ninety", all contain either "e" or "o".
This covers all natural numbers, as desired.