Define $a_k = (k^2 + 1)k!$ and $b_k = a_1 + a_2 + a_3 + \cdots + a_k$. Let \[\frac{a_{100}}{b_{100}} = \frac{m}{n}\] where $m$ and $n$ are relatively prime natural numbers. Find $n - m$.
1. Define the sequences \(a_k\) and \(b_k\) as follows:
   \[
   a_k = (k^2 + 1)k!
   \]
   \[
   b_k = a_1 + a_2 + a_3 + \cdots + a_k
   \]

2. We start by simplifying \(a_k\):
   \[
   a_k = (k^2 + 1)k! = k^2 k! + k! = k(k \cdot k!) + k! = k(k+1)! - k \cdot k! + k!
   \]
   \[
   = k(k+1)! - k!(k-1)
   \]

3. Next, we express \(b_k\) in terms of the sum of \(a_i\):
   \[
   b_k = a_1 + a_2 + a_3 + \cdots + a_k
   \]
   Using the expression for \(a_k\) derived above:
   \[
   b_k = \sum_{i=1}^k \left[ i(i+1)! - i!(i-1) \right]
   \]

4. Notice that this sum is telescoping:
   \[
   b_k = \left[ 1 \cdot 2! - 1! \cdot 0 \right] + \left[ 2 \cdot 3! - 2! \cdot 1 \right] + \left[ 3 \cdot 4! - 3! \cdot 2 \right] + \cdots + \left[ k \cdot (k+1)! - k! \cdot (k-1) \right]
   \]
   Most terms cancel out, leaving:
   \[
   b_k = k(k+1)!
   \]

5. Now, we need to find \(\frac{a_{100}}{b_{100}}\):
   \[
   \frac{a_{100}}{b_{100}} = \frac{(100^2 + 1)100!}{100 \cdot 101!}
   \]
   Simplify the fraction:
   \[
   \frac{a_{100}}{b_{100}} = \frac{100^2 + 1}{100 \cdot 101} = \frac{10000 + 1}{100 \cdot 101} = \frac{10001}{10100}
   \]

6. We need to ensure that the fraction \(\frac{10001}{10100}\) is in its simplest form. Since 10001 and 10100 are relatively prime (they have no common factors other than 1), the fraction is already simplified.

7. Let \(\frac{10001}{10100} = \frac{m}{n}\), where \(m = 10001\) and \(n = 10100\). We are asked to find \(n - m\):
   \[
   n - m = 10100 - 10001 = 99
   \]

The final answer is \(\boxed{99}\)