Well, there must be an almost-infinite number of possible solutions, so I'll post my one (which is probably far from intended answer, but still not too far-fetched to match this exact question, I believe).
- List 1 contains all even numbers, which either have an even number of prime divisors, or have at least one prime divisor containing digit 3 (in decimal expansion): 10, 20, 34, 72, 80 and 96 all have 2 prime divisors each (2 and 5, 2 and 5, 2 and 17, 2 and 3, 2 and 5, 2 and 3 respectively), while 130 and 132, despite having 3 prime divisors (2,5,13 and 2,3,11 respectively), can be divided by a prime which has a 3 in it (3 and 13, respectively).
- On the other hand, List 2 contains all other numbers (i.e. all odd numbers and even numbers with an odd count of prime divisors, neither of them containing a 3), e.g. 7, 11, 59, 81, 99 are all odd, while 32 ($=2^5$), 140 ($=2^2\times5\times7$) and 190 ($=2\times5\times19$) contain an odd number of prime divisors, neither of them has a 3 in its decimal expansion.
Note that
there are more possible numbers for List 2 than for List 1 (from a set of positive integers not greater than some given number $N$, i.e. $\{1,2,\dots,N\}$), because the latter contains only part of the even numbers, while the former contains all odd and some even ones (and the count of even numbers in $\{1,2,\dots,N\}$ is never greater than the count of odd ones).
So
22 ($=2\times11$ - 2 prime divisors), 150 ($=2\times3\times5^2$ - 3 prime divisors but including the 3) and 160 ($=2^5\times5$ - 2 prime divisors) go into List 1, while all other numbers - 5, 29, 33, 83 and 107 (being all odd) - go into List 2.
