Let $ k $ be the number of tiles. There are two cases: If $ k $ has twenty divisors, then we can divide them into ten pairs, which gives us 10 ways to write $ k $ as the product of two positive integers. Alternatively, if $ k $ has 19 divisors, then $ k $ is a square. So other than the square case, there are $ (19 - 1)/2 = 9 $ ways to write $ k $ as the product of two positive integers, which gives us a total of $ 9 + 1 = 10 $ ways. If the prime factorization of $ k $ is $ p_1^{e_1} p_2^{e_2} \dotsm p_n^{e_n}, $ then the number of divisors of $ k $ is \[(e_1 + 1)(e_2 + 1) \dotsm (e_n + 1).\]Note that $ e_i \ge 1 $ for each $ i, $ so each factor $ e_i + 1 $ is at least 2. If $ k $ has 19 divisors, then $ k $ must be of the form $ p^{18}, $ where $ p $ is prime. The smallest number of this form is $ 2^{18} = 262144 $. Otherwise, $ k $ has 20 divisors. We want to write 20 as the product of factors, each of which are least 2. Here are all the ways: \[20 = 2 \cdot 10 = 4 \cdot 5 = 2 \cdot 2 \cdot 5.\] Thus, we have the following cases. (i) $ k=p^{19} $ for some prime $ p $. The smallest such $ k $ is attained when $ p=2, $ which gives $ k=2^{19} $. (ii) $ k=pq^9 $ for distinct primes $ p $ and $ q $. The smallest such $ k $ is attained when $ p = 3 $ and $ q = 2 $ which gives $ k=2^9\cdot3 $. (iii) $k=p^3 q^4 $ for distinct primes $ p $ and $ q $. The smallest such $ k $ is attained when $ p = 3 $ and $ q = 2, $ which gives $ k=2^4\cdot3^3=432 $. (iv) $k=pqr^4 $ for distinct primes $ p, $ $ q, $ and $ r $. The smallest such $ k $ is attained when $ p = 3, $ $ q = 5, $ and $ r = 2, $ which gives $ k=2^4\cdot3\cdot5=240 $. Therefore, the least number of tiles Emma could have is $ \boxed{240} $ tiles.