Let $\Phi$ and $\Psi$ denote the Euler totient and Dedekind‘s totient respectively. Determine all $n$ such that $\Phi(n)$ divides $n +\Psi (n)$.
1. **Identify the problem and given conditions:**
   We need to determine all \( n \) such that \(\Phi(n)\) divides \( n + \Psi(n) \), where \(\Phi\) is the Euler totient function and \(\Psi\) is the Dedekind's totient function.

2. **Initial solution set:**
   The solution set is given as \(\{1, 2^{n_1}, 2^{n_1}3^{n_2}, 2^{n_1}5^{n_2} : n_1, n_2 \in \mathbb{N}\}\).

3. **Base case \( n = 1 \):**
   It's clear that \( n = 1 \) is a solution because \(\Phi(1) = 1\) and \(\Psi(1) = 1\), so \( \Phi(1) \mid 1 + \Psi(1) \) holds true.

4. **General case \( n > 1 \):**
   Suppose \( n > 1 \) with prime factorization \( n = \prod_{i=1}^k p_i^{n_i} \).

5. **Expression for \(\Phi(n)\) and \(\Psi(n)\):**
   \[
   \Phi(n) = n \prod_{i=1}^k \left(1 - \frac{1}{p_i}\right)
   \]
   \[
   \Psi(n) = n \prod_{i=1}^k \left(1 + \frac{1}{p_i}\right)
   \]

6. **Condition \(\Phi(n) \mid n + \Psi(n)\):**
   This implies:
   \[
   \Phi(n) \mid n + \Psi(n) \implies n \prod_{i=1}^k \left(1 - \frac{1}{p_i}\right) \mid n + n \prod_{i=1}^k \left(1 + \frac{1}{p_i}\right)
   \]
   Simplifying, we get:
   \[
   n \prod_{i=1}^k \left(1 - \frac{1}{p_i}\right) \mid n \left(1 + \prod_{i=1}^k \left(1 + \frac{1}{p_i}\right)\right)
   \]
   Dividing both sides by \( n \):
   \[
   \prod_{i=1}^k \left(1 - \frac{1}{p_i}\right) \mid 1 + \prod_{i=1}^k \left(1 + \frac{1}{p_i}\right)
   \]

7. **Simplifying the divisibility condition:**
   Let \( m \) be an integer such that:
   \[
   m \prod_{i=1}^k (p_i - 1) = \prod_{i=1}^k p_i + \prod_{i=1}^k (p_i + 1)
   \]
   This is equivalent to:
   \[
   m \prod_{i=1}^k (p_i - 1) = \prod_{i=1}^k p_i + \prod_{i=1}^k (p_i + 1)
   \]

8. **Considering the parity of the terms:**
   If \( p_i \neq 2 \) for all \( i \), then the left-hand side (LHS) is even, but the right-hand side (RHS) is odd, which is not possible. Therefore, we must have \( p_1 = 2 \).

9. **Divisibility by \( 2^{k-1} \):**
   Since \( 2^{k-1} \) divides both \( \prod_{i=1}^k (p_i - 1) \) and \( \prod_{i=1}^k (p_i + 1) \), it must also divide \( \prod_{i=1}^k p_i \). This implies \( k = 1 \) or \( k = 2 \) because \( p_2, \ldots, p_k \) are all odd.

10. **Case \( k = 1 \):**
    If \( k = 1 \), then \( n = 2^{n_1} \) and any such \( n \) satisfies the condition.

11. **Case \( k = 2 \):**
    If \( k = 2 \), then \( n = 2^{n_1} p_2^{n_2} \). By the condition:
    \[
    m (p_2 - 1) = 2 p_2 + 3 (p_2 + 1)
    \]
    Simplifying, we get:
    \[
    m = \frac{5 p_2 + 3}{p_2 - 1} = 5 + \frac{8}{p_2 - 1}
    \]
    This implies \( p_2 - 1 \in \{1, 2, 4, 8\} \). Since \( p_2 \) is an odd prime, we get \( p_2 = 3 \) or \( p_2 = 5 \). Thus, \( n = 2^{n_1} 3^{n_2} \) or \( n = 2^{n_1} 5^{n_2} \).

The final answer is \( \boxed{ \{1, 2^{n_1}, 2^{n_1}3^{n_2}, 2^{n_1}5^{n_2} : n_1, n_2 \in \mathbb{N}\} } \).