We will find the positive divisors of 14 by finding pairs that multiply to 14. We begin our list as follows, $$1 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 14.$$ Checking $ 2 $, we find that $ 2\cdot 7=14 $, so our list becomes $$1 \quad 2 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 7 \quad 14.$$ Checking $ 3 $, $ 4 $, $ 5 $, and $ 6 $, we find that none of these are divisors of $ 14 $, so our final list is $$1 \quad 2 \quad 7 \quad 14.$$ Next, we use the buddy method to determine the factors of $ 42 $. We begin our list as follows, $$1\quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 42.$$ Checking $ 2 $, we find that $ 2\cdot 21=42 $, so our list becomes $$1\quad 2 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 21 \quad 42.$$ Checking $ 3 $, we find that $ 3\cdot 14=42 $, so our list becomes $$1\quad 2 \quad 3 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 14 \quad 21 \quad 42.$$ Checking $ 4 $ and $ 5 $ we find that $ 4 $ and $ 5 $ are not divisors of $ 42 $. Checking $ 6 $, we find that $ 6\cdot 7=42 $, so our list becomes $$1\quad 2 \quad 3 \quad 6 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 7 \quad 14 \quad 21 \quad 42.$$ Since $ 7 $ is already on our list, our final list is $$1\quad 2 \quad 3 \quad 6 \quad 7 \quad 14 \quad 21 \quad 42.$$ We compare our lists for the factors of $ 14 $ and the factors of $ 42 $ to see that the factors that $ 14 $ and $ 42 $ share are $ 1 $, $ 2 $, $ 7 $, and $ 14 $. Therefore, Rick and Steve could be thinking of $ \boxed{4} $ possible numbers. Note that since $ 14 $ is a factor of $ 42 $, all of the factors of $ 14 $ are also factors of $ 42 $.