jmc

$$252=2^2\cdot 3^2\cdot 7^1$$An even number contains at least one power of 2 in its prime factorization. This means that an even divisor of 252 must be in the form $2^a\cdot 3^b\cdot 7^c$, where there are 2 choices for $a$ (1 or 2), three are 3 choices for $b$ (0, 1, or 2), and 2 choices for $c$ (0 or 1). This means that $2\cdot 3\cdot 2=\boxed{12}$ of the positive divisors of 252 are even. Now, see if you can find a complementary counting approach.