Define the sequences $(a_n)$ and $(b_n)$ by
$$\begin{align*} & a_n, b_n > 0, \forall n \in \mathbb{N_+}, \\ & a_{n+1} = a_n - \frac{1}{1 + \sum_{i=1}^n \frac{1}{a_i}}, \\ & b_{n+1} = b_n + \frac{1}{1 + \sum_{i=1}^n \frac{1}{b_i}}. \end{align*}$$

1. If $a_{100} b_{100} = a_{101} b_{101}$, find the value of $a_1 - b_1$.

First, we derive the relationship for $a_n$:
$$a_{n+1} \left( 1 + \sum_{i=1}^n \frac{1}{a_i} \right) = a_n \left( 1 + \sum_{i=1}^n \frac{1}{a_i} \right) - 1.$$
Iterating this, we get:
$$a_{n+1} \left( 1 + \sum_{i=1}^n \frac{1}{a_i} \right) = a_1.$$
Thus,
$$\frac{a_1}{a_{n+1}} = 1 + \sum_{i=1}^n \frac{1}{a_i}.$$
For $b_n$, we have:
$$b_{n+1} \left( 1 + \sum_{i=1}^n \frac{1}{b_i} \right) = b_n \left( 1 + \sum_{i=1}^n \frac{1}{b_i} \right) + 1.$$
Iterating this, we get:
$$b_{n+1} \left( 1 + \sum_{i=1}^n \frac{1}{b_i} \right) = b_1 + 2n.$$
Thus,
$$\frac{b_1 + 2n}{b_{n+1}} = 1 + \sum_{i=1}^n \frac{1}{b_i}.$$

Given $a_{100} b_{100} = a_{101} b_{101}$, we have:
$$\frac{a_1}{a_{101}} = \frac{a_1}{a_{100}} \cdot \frac{a_{100}}{a_{101}} = 1 + \sum_{i=1}^{100} \frac{1}{a_i},$$
and
$$\frac{b_1 + 198}{b_{100}} = \frac{b_1 + 198}{b_{101}} \cdot \frac{b_{101}}{b_{100}} = 1 + \sum_{i=1}^{99} \frac{1}{b_i}.$$

Equating the expressions, we get:
$$\frac{a_1}{a_{101}} = \frac{b_1 + 198}{b_{100}}.$$
Given $a_{100} b_{100} = a_{101} b_{101}$, we have:
$$\frac{a_{101}}{a_{100}} = \frac{b_{100}}{b_{101}}.$$
Thus,
$$a_1 - b_1 = 199.$$

The answer is: \boxed{199}.

2. If $a_{100} = b_{99}$, determine which is larger between $a_{100} + b_{100}$ and $a_{101} + b_{101}$.

We need to compare $a_{100} + b_{100}$ and $a_{101} + b_{101}$. We have:
$$a_{101} = a_{100} - \frac{1}{1 + \sum_{i=1}^{100} \frac{1}{a_i}},$$
and
$$b_{101} = b_{100} + \frac{1}{1 + \sum_{i=1}^{100} \frac{1}{b_i}}.$$

Since $a_{100} = b_{99}$, we need to analyze the behavior of the sequences. Given the nature of the sequences, $a_{n+1}$ is always less than $a_n$ and $b_{n+1}$ is always greater than $b_n$.

Thus, $a_{101} < a_{100}$ and $b_{101} > b_{100}$. Therefore:
$$a_{100} + b_{100} > a_{101} + b_{101}.$$

The answer is: $a_{100} + b_{100}$ is larger.