(1) Set $a_1 = a$ and $b_1 = b$ ($a, b > 0$). The recurrence formula of $a_n$ implies that, for any positive integer $n \ge 2$, we have
$$\frac{1}{a_n - a_{n+1}} = 1 + \sum_{i=1}^{n} \frac{1}{a_i} \implies \frac{1}{a_n - a_{n+1}} = \frac{1}{a_{n-1} - a_n} + \frac{1}{a_n}.$$
That is, $\frac{a_n}{a_n - a_{n-1}} = \frac{a_n}{a_{n-1} - a_n} + 1 = \frac{a_{n-1}}{a_{n-1} - a_n}$. From this, we obtain $\frac{a_{n+1}}{a_n} = \frac{a_n}{a_{n-1}}$.
So $\{a_n\}$ is a geometric sequence with common ratio $\frac{a}{a+1}$, and the general term is
$$a_n = \frac{a^n}{(a+1)^{n-1}}.$$
Similarly, the recurrence formula for $b_n$ implies that
$$\frac{1}{b_{n+1} - b_n} = 1 + \sum_{i=1}^{n} \frac{1}{b_i} = \frac{1}{b_n - b_{n-1}} + \frac{1}{b_n}.$$
Thus, $\frac{b_n}{b_{n+1} - b_n} = \frac{b_n}{b_{n-1} - b_n} + 1$. By induction, we have $\frac{b_n}{b_{n+1} - b_n} = \frac{b_1}{b_2 - b_1} + 2(n-1) = b + 2n - 1$. Therefore, $\frac{b_{n+1}}{b_n} = \frac{b + 2n}{b + 2n - 1}$.
From this, we get $b_n = b \prod_{k=1}^{n-1} \frac{b+2k}{b+2k-1}$.
The condition $a_{100}b_{100} = a_{101}b_{101}$ in (1) implies that $\frac{a_{100}}{a_{101}} = \frac{b_{101}}{b_{100}}$, i.e. $\frac{a+1}{a} = \frac{b+200}{b+199}$. So $a = b + 199$ and thus $a_1 - b_1 = a - b = 199$.

(2) Method 1. It is clear that $\{a_n\}$ is monotonically decreasing and $\{b_n\}$ is monotonically increasing. Together with the condition $a_{100} = b_{99}$, we deduce that
(*) $a_1 > a_2 > \dots > a_{99} > a_{100} = b_{99} > b_{98} > \dots > b_1 > 0$.
In turn, we deduce that
$$a_{99} = a_{100} + \frac{1}{1 + \sum_{i=1}^{99} \frac{1}{a_i}} > b_{99} + \frac{1}{1 + \sum_{i=1}^{99} \frac{1}{b_i}} = b_{100}.$$

Method 2. Since $a_{100} = b_{99}$, we have
$$(**) \qquad \frac{a^{100}}{(a+1)^{99}} = \frac{b(b+2)\cdots(b+196)}{(b+1)(b+3)\cdots(b+195)}.$$
To prove $a_{100} + b_{100} > a_{101} + b_{101}$, it is equivalent to prove that
$$\begin{aligned} a_{100} - a_{101} &> b_{101} - b_{100} \\ \Longleftrightarrow \frac{a_{100} - a_{101}}{a_{100}} &> \frac{b_{101} - b_{100}}{b_{99}} \\ \Longleftrightarrow 1 - \frac{a}{a+1} &> \frac{(b+198)(b+200)}{(b+197)(b+199)} - \frac{b+198}{b+197} \\ \Longleftrightarrow \frac{1}{a+1} &> \frac{b+198}{(b+197)(b+199)} \\ (*) & \Longleftrightarrow a < b+197 - \frac{1}{b+198}. \end{aligned}$$
We prove this by contradiction. Suppose that $a \ge b + 197 - \frac{1}{b+198}$, in particular, $a > b + 196$. We have
$$\begin{aligned} \frac{a^2}{a+1} - (b+196) &= \frac{a(a-b-196) - (b+196)}{a+1} \\ &\ge \frac{1}{a+1} \left( \left( b+197 - \frac{1}{b+198} \right) \left( 1 - \frac{1}{b+198} \right) - (b+196) \right) \\ &> \frac{1}{a+1} \left( \left( b+197 - \frac{1}{b+197} \right) \frac{b+197}{b+198} - (b+196) \right) = 0. \end{aligned}$$
But note that $\frac{a}{a+1} > \frac{b+194}{b+195} > \frac{b+192}{b+193} > \cdots > \frac{b}{b+1}$, we have
$$\frac{a^{100}}{(a+1)^{99}} = \left(\frac{a}{a+1}\right)^{98} \frac{a^2}{a+1} > \frac{b}{b+1} \cdots \frac{b+194}{b+195} (b+196),$$
contradicting with ()! So (*) holds, and thus $a_{100} + b_{100} > a_{101} + b_{101}$.