For each positive integer $n$, write the sum $\sum_{m=1}^n 1/m$ in the form $p_n/q_n$, where $p_n$ and $q_n$ are relatively prime positive integers.  Determine all $n$ such that 5 does not divide $q_n$.
To solve the problem, we need to determine all positive integers \( n \) such that the denominator \( q_n \) of the sum \( \sum_{m=1}^n \frac{1}{m} \) in its lowest terms is not divisible by 5. We will use the given theorems and corollaries to systematically identify these integers.

1. **Theorem I**: If \( \frac{a_i}{b_i} \) are fractions in their lowest terms for \( i=1, 2, 3 \), \( \frac{a_1}{b_1} + \frac{a_2}{b_2} = \frac{a_3}{b_3} \), and \( b_1 \) is not divisible by 5, then the highest power of 5 dividing \( b_2 \) and \( b_3 \) are the same.

   **Proof**: 
   - If 5 is not a factor of \( b_2 \), then \( b_3 \) is a factor of \( b_1 b_2 \) and hence not divisible by 5.
   - If \( b_2 \) is a multiple of 5, then \( a_2 \) is not, and hence \( a_2 b_1 \) is not, and hence \( a_2 b_1 + a_1 b_2 \) is not a multiple of 5.
   - Thus, since \( b_3 = \frac{b_1 b_2}{\gcd(b_1 b_2, a_2 b_1 + a_1 b_2)} \), \( b_3 \) and \( b_2 \) are divisible by the same powers of 5.

2. **Corollary II**: For \( 0 \le k \le 4 \) and positive integer \( n \), the highest power of 5 dividing the denominator of \( \frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{5n+k} \) is the same as that dividing the denominator of \( \frac{1}{5} + \frac{1}{10} + \ldots + \frac{1}{5n} \).

   **Proof**: The integer reciprocals omitted in the latter sum give a fraction whose denominator is not divisible by 5.

3. **Corollary III**: For \( 0 \le k \le 4 \) and positive integer \( n \), \( q_{5n+k} \) is not divisible by 5 if and only if \( p_n \) is divisible by 5.

   **Proof**: Corollary II states that the highest power of 5 dividing \( q_{5n+k} \) is the same as the highest power of five dividing the lowest-terms denominator of \( \frac{1}{5} + \frac{1}{10} + \ldots + \frac{1}{5n} = \frac{p_n}{5q_n} \). This will be a multiple of 5 if \( p_n \) is not divisible by 5. If, on the other hand, \( p_n \) is a multiple of 5, then \( q_n \) is not, so the 5 will cancel.

4. **Corollary IV**: If we define \( D \) to be the set of integers such that \( q_n \) is not divisible by 5 and \( N \) to be the set of integers such that \( p_n \) is divisible by 5, then \( N \) is a subset of \( D \) and for \( 0 \le k \le 4 \) and \( n > 0 \), \( 5n+k \in D \) iff \( n \in N \).

   **Proof**: \( 5n+k \in D \) iff \( n \in N \) is the content of Corollary III. If we let \( n \) be some element of \( N \) that is not in \( D \), then the coprime integers \( p_n \) and \( q_n \) are both divisible by 5, which is a contradiction.

5. **Lemma V**: \( \{1,2,3\} \subset (D \setminus N) \), while \( 4 \in N \subset D \).

   **Proof**: Let us consider the integers 1 to 4 separately:
   - \( \frac{1}{1} = \frac{1}{1} \), so 1 is in \( D \) but not \( N \).
   - \( \frac{1}{1} + \frac{1}{2} = \frac{3}{2} \), so 2 is in \( D \) but not \( N \).
   - \( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{11}{6} \), so 3 is in \( D \) but not \( N \).
   - \( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{25}{12} \), so 4 is in \( D \) and in \( N \).

6. **Theorem VI**: \( N = \{4, 20, 24\} \), so \( D = \{1, 2, 3, 4, 20, 21, 22, 23, 24, 100, 101, 102, 103, 104, 120, 121, 122, 123, 124\} \).

   **Partial Proof**:
   - Since 4 is in \( N \), it follows that the numbers 20 up to 24 are in \( D \).
   - By inspection, 20 and 24 are the numbers in this range in \( N \).
   - Thus the numbers in the ranges \( [100..104] \) and \( [120..124] \) are both contained in \( D \).
   - By further inspection, none of these numbers are in \( N \).

   Then given a minimal number \( n \) other than \( \{4, 20, 24\} \) in \( N \), \( n \ge 5 \) by Lemma V. Then \( n = 5k + r \) for some positive integer \( k \) and integer \( 0 \le r \le 4 \). Then \( n \) is in \( N \) which means that \( n \) is in \( D \) and hence \( k \) is in \( N \) by Corollary IV. By the minimality of \( N \), \( k \in \{4, 20, 24\} \) and hence \( n \in \{21, 22, 23, 100, 101, 102, 103, 104, 120, 121, 122, 123, 124\} \). However, we have already inspected that none of these \( n \) are in \( N \).

The final answer is \( \boxed{ \{1, 2, 3, 4, 20, 21, 22, 23, 24, 100, 101, 102, 103, 104, 120, 121, 122, 123, 124\} } \).