First, any number $ n$ is abundant if $ \sigma(n)>2n$. Also, if the prime factorization of $ n$ is $ p_1^{e_1}p_2^{e_2}...p_{m}^{e_m}$, then it is well known that $ \sigma(n)=\prod_{i=1}^{m}(1+p_i+\cdots+p_i^{e_i})$. First, I claim that any multiple $ ab$ of an abundant number $ a$ is abundant. If $ d_1,...d_k$ are the divisors, of $ a$, then \begin{align*} \sigma(a)=d_1+d_2+\cdots +d_k&>2a\\ bd_1+bd_2+\cdots +bd_k&>2ab \end{align*} and the $ bd_1,bd_2,...,bd_k$ is only a subset of the divisors of $ ab$, so $ \sigma(ab)>2ab$. It can be checked that $ 88=2^3\cdot 11$ is abundant, because $ \sigma(88)=(1+2+4+8)(1+11)=15\cdot 12=180>2\cdot 88$. Thus any multiple of $ 88$ is abundant. Now I claim that any number of the form $ 315a=3^2\cdot 5\cdot 7\cdot a$ is abundant if $ 2\le a\le 89$. It suffices to show that it works if $ a$ is any prime less than or equal to $ 89$. We'll test this individually for $ a=3,5,7$. \[ \sigma(945)=\sigma(3^3\cdot 5\cdot 7)=(1+3+9+27)(1+5)(1+7)=40\cdot 6\cdot 8=1920>2\cdot 945\] \[ \sigma(1575)=\sigma(3^2\cdot 5^2\cdot 7)=(1+3+9)(1+5+25)(1+7)=13\cdot 31\cdot 8=3224>2\cdot 1575\] \[ \sigma(2205)=\sigma(3^2\cdot 5\cdot 7^2)=(1+3+9)(1+5)(1+7+49)=13\cdot 6\cdot 57=4446>2\cdot 2205\] Now assume $ a$ is a prime less than or equal to 89. Then $ \frac{\sigma(a)}{a}=\frac{a+1}{a}=1+\frac{1}{a}\ge 1+\frac{1}{89}=\frac{90}{89}$. Thus \begin{align*} \frac{\sigma(315a)}{315a}&=\frac{\sigma(315)\sigma(a)}{315a}\\ &\ge\frac{\sigma(315)}{315}\cdot\frac{90}{89}\\ &=\frac{(1+3+9)(1+5)(1+7)}{315}\cdot\frac{90}{89}\\ &=\frac{56160}{28035}\\ &>2 \end{align*}. It is well know that any number greater than or equal to $ 88\cdot 315-88-315$ is expressible as the sum $ 88A+315B$ where $ A\ge 0$ and $ B\ge 0$. We can also ensure that $ B\le 87$. If $ B\ge 88$, then we can subtract $ 88$ from $ B$ and add $ 315$ to $ A$, and continue this until we have $ B\le 87$. Therefore, any number greater than $ 88\cdot 315-88-315+(2\cdot 315+88)=89\cdot 315$ can be expressed as $ (A+1)88+(B+2)315$ with have $ 2\le B+2\le 89$. Then $ (A+1)88$ and $ (B+2)315$ are both abundant numbers, and we are done.