1. **Constructing the sequence $a_1, a_2, \ldots, a_{101}$:**
- We need to ensure that for any non-empty subset $S$ of $\{1, 2, \ldots, 101\}$, the sum $\sum_{i \in S} a_i$ divides $100! + \sum_{i \in S} b_i$.
- We impose the condition $a_i \equiv 1 \pmod{(100!)!}$ for all $i$. This ensures that $a_i$ are congruent to 1 modulo $(100!)!$.

2. **Ensuring distinct sums modulo primes $p \geq 101$:**
- We need to ensure that no two different sets $S_1$ and $S_2$ have sums $\sum_{i \in S_1} a_i$ and $\sum_{i \in S_2} a_i$ that are divisible by the same prime $p \geq 101$.
- This can be achieved by choosing $a_i$ such that no sum $\sum_{i \in S} a_i$ is divisible by any prime $p$ in the range $101 \leq p \leq (100!)!$, except for the sum of all $a_i$, which is divisible by 101.

3. **Inductive construction of $a_i$:**
- Suppose $a_1, a_2, \ldots, a_k$ have been chosen to satisfy the condition.
- We need to choose $a_{k+1}$ such that no prime $p > (100!)!$ divides two different sums, at least one of which contains $a_{k+1}$.
- If $p$ divides $\sum_{i \in S_1} a_i - \sum_{i \in S_2} a_i$ for some $S_1, S_2 \subseteq \{1, 2, \ldots, k\}$, we call $p$ a bad prime.
- The set of bad primes is finite, and by the Chinese Remainder Theorem (CRT), we can choose $a_{k+1}$ such that no sum containing $a_{k+1}$ is divisible by a bad prime.

4. **Using CRT to choose $a_{k+1}$:**
- By the CRT, we can find $a_{k+1}$ such that it satisfies the required conditions modulo all bad primes.
- This process can be repeated inductively to construct $a_1, a_2, \ldots, a_{101}$.

5. **Constructing $b_1, b_2, \ldots, b_{101}$:**
- After constructing $a_1, a_2, \ldots, a_{101}$, we can construct $b_1, b_2, \ldots, b_{101}$ using the CRT.
- We ensure that $100! \mid b_i$ for all $i$, and for each prime $p > 100$, $p$ divides at most one sum $\sum_{i \in S} a_i$.