What is the smallest $n$ such that the condition holds? What is the smallest $n$ such that for all $a,b$ with $12 \mid a+b$, it is the case that if $n\mid ab$, then $12 \mid a$ and $12 \mid b$?

I think I see how it works if $n = 144$, but I don't know how to prove that it is the smallest (or that there is another value of $n$ which is).
Let's see. Apparently, $n$ is divisible by 2 and 3, otherwise it couldn't enforce divisibility by 12. Now, the product $ab$ is divisible by 3, which means that at least one of the numbers $a$ and $b$ is divisible by 3, and so is their sum, hence so is the other number. By similar reasoning, both are divisible by 2. So $a=6x,\;b=6y$. Now the question changes to: given that $x+y$ is even, what is the condition on $n$, such that ${n\over36}\mid xy$, that would make both $x$ and $y$ even? The answer is obvious: the product must be even as well, so $n=72$ will suffice.
If I get you right, we have to find $n,a,b$ so that the conditions 


$12 \mid a+b $
$n \mid ab $
$12 \mid a $
$12 \mid b $


hold true.

Well, if you are searching for the smallest possible values, than you might be able to simply check the first numbers with a simple, however inefficient, computer program. This is a Python example:

import itertools
for i in itertools.product(range(1,100),range(1,100),range(1,100)):
    n,a,b = i
    if ((a + b) % 12 == 0) and ((a * b) % n == 0):
        if (a % 12 == 0) and (b % 12 == 0):
            print n,a,b


The results are:


1 12 12
1 12 24
1 12 36
1 12 48
1 12 60
1 12 72
...