Let $a_1,a_2,...,a_m(m\geq 1)$ be all the positive divisors of $n$. If there exist $m$ integers $b_1,b_2,...b_m$ such that $n=\sum_{i=1}^m (-1)^{b_i} a_i$, then $n$ is a $\textit{good}$ number. Prove that there exist a good number with exactly $2013$ distinct prime factors.
I'll prove a stronger statement: there exists a good number with exactly $k$ distinct prime factors, iff $k\ge 2$.
It's easy to see that $k=1$ does not work, as taking mod $p$ gives $0$ on one side and $\pm 1$ on the other, contradiction.
The construction for $k\ge 2$ is as follows: Define $n=p_1p_2\cdots p_k$ where $p_i$ are all distinct. Let $p_1=2, p_2=3$. Then all factors of $n$ can be described as: $1\cdot n', 2\cdot n', 3\cdot n', \text{ or }6\cdot n'$ for $n'\mid p_3\cdots p_k$.
If $n'\ne p_3\cdots p_k$, then add the factors as follows: $$1\cdot n' + 2\cdot n' + 3\cdot n' - 6\cdot n'=0$$If $n' = p_3\cdots p_k$, then add the factors as follows: $$1\cdot n'+2\cdot n'-3\cdot n' + 6\cdot n'=6n' = n$$which means $n$ is a good number. Setting $k=2013$ solves the problem at hand.
Extension: Prove that there exists a suitable good number $n$ for any $m$ that we choose iff $m$ is not prime.