For a positive integer $n$ that is not a power of two, we define $t(n)$ as the greatest odd divisor of $n$ and $r(n)$ as the smallest positive odd divisor of $n$ unequal to $1$. Determine all positive integers $n$ that are not a power of two and for which we have $n = 3t(n) + 5r(n)$.
To solve the problem, we need to determine all positive integers \( n \) that are not a power of two and satisfy the equation \( n = 3t(n) + 5r(n) \), where \( t(n) \) is the greatest odd divisor of \( n \) and \( r(n) \) is the smallest positive odd divisor of \( n \) unequal to 1.

1. **Express \( n \) in terms of its odd and even components:**
   Let \( n = s \cdot 2^r \), where \( s \) is an odd integer and \( r \geq 1 \). Since \( s \) is odd, \( t(n) = s \).

2. **Analyze the given equation:**
   Substitute \( t(n) = s \) into the equation:
   \[
   n = 3t(n) + 5r(n) \implies s \cdot 2^r = 3s + 5r(n)
   \]
   Rearrange to isolate \( r(n) \):
   \[
   s \cdot 2^r - 3s = 5r(n) \implies s(2^r - 3) = 5r(n)
   \]
   Since \( r(n) \) is the smallest positive odd divisor of \( n \) unequal to 1, \( r(n) \) must be a divisor of \( s(2^r - 3) \).

3. **Consider the possible values of \( r \):**
   - For \( r = 2 \):
     \[
     s(2^2 - 3) = s \cdot 1 = 5r(n) \implies s = 5r(n)
     \]
     Since \( r(n) \) is the smallest positive odd divisor of \( n \) unequal to 1, \( r(n) \) must be a divisor of \( s \). Therefore, \( s = 5 \) and \( r(n) = 1 \), which contradicts the condition that \( r(n) \neq 1 \). Hence, \( r \neq 2 \).

   - For \( r = 3 \):
     \[
     s(2^3 - 3) = s \cdot 5 = 5r(n) \implies s = r(n)
     \]
     Here, \( s \) must be an odd prime number because \( r(n) \) is the smallest positive odd divisor of \( n \) unequal to 1. Therefore, \( n = 8s \), where \( s \) is an odd prime.

4. **Verify the solutions:**
   - For \( n = 8p \) where \( p \) is an odd prime:
     \[
     t(n) = p \quad \text{and} \quad r(n) = p
     \]
     Substitute into the original equation:
     \[
     n = 3t(n) + 5r(n) \implies 8p = 3p + 5p \implies 8p = 8p
     \]
     This holds true, so all numbers of the form \( 8p \) where \( p \) is an odd prime are solutions.

5. **Consider other possible values of \( s \):**
   - For \( r = 2 \):
     \[
     s(2^2 - 3) = s \cdot 1 = 5r(n) \implies s = 5r(n)
     \]
     Since \( r(n) \) must be a divisor of \( s \), \( s \) must be of the form \( 5k \) where \( k \) is an odd integer. The smallest such \( s \) are \( 15 \) and \( 25 \).

   - For \( s = 15 \):
     \[
     n = 15 \cdot 2^2 = 60
     \]
     Verify:
     \[
     t(60) = 15 \quad \text{and} \quad r(60) = 3
     \]
     Substitute into the original equation:
     \[
     60 = 3 \cdot 15 + 5 \cdot 3 \implies 60 = 45 + 15 \implies 60 = 60
     \]
     This holds true.

   - For \( s = 25 \):
     \[
     n = 25 \cdot 2^2 = 100
     \]
     Verify:
     \[
     t(100) = 25 \quad \text{and} \quad r(100) = 5
     \]
     Substitute into the original equation:
     \[
     100 = 3 \cdot 25 + 5 \cdot 5 \implies 100 = 75 + 25 \implies 100 = 100
     \]
     This holds true.

Conclusion:
The solutions are \( n = 8p \) where \( p \) is an odd prime, \( n = 60 \), and \( n = 100 \).

The final answer is \( \boxed{8p, 60, 100} \) where \( p \) is an odd prime.