We say that a positive integer $n$ is excellent if it is good, and we can find integers $c_1, c_2, \dots, c_m$ such that $\sum_{j=1}^{m} (-1)^{c_j} a_j = 0$. Let $p_1, p_2, \dots, p_k$ be distinct primes larger than 3. We prove by induction on $k$ that the integer $6p_1p_2\cdots p_k$ is excellent.

For the base case $k=0$, the positive divisors of $6$ are $1$, $2$, $3$, $6$. Since we have $-1-2+3+6=6$ and $-1-2-3+6=0$, the integer $6$ is excellent.

For the inductive step, assume $n = 6p_1p_2\cdots p_k$ is excellent. Let $a_1, a_2, \dots, a_m$ be all of its positive divisors, and let
$$\sum_{j=1}^{m} (-1)^{b_j} a_j = n \quad \text{and} \quad \sum_{j=1}^{m} (-1)^{c_j} a_j = 0.$$
Consider $n' = 6p_1p_2\cdots p_{k+1}$. Its positive divisors are of the form $a_j$ and $a_jp_{k+1}$. Note that
$$\sum_{j=1}^{m} (-1)^{c_j} a_j + \sum_{j=1}^{m} (-1)^{b_j} (a_j p_{k+1}) = 0 + n p_{k+1} = n'.$$
Also, we have
$$\sum_{j=1}^{m} (-1)^{c_j} a_j + \sum_{j=1}^{m} (-1)^{b_j} (a_j p_{k+1}) = 0 + 0 \cdot p_{k+1} = 0.$$
Therefore, $n'$ is excellent. By induction, our claim holds. In particular, we can take $k=2011$ so that $6p_1p_2\cdots p_{2011}$ is excellent, and hence it is a good integer with exactly 2013 prime divisors.