Find the smallest possible number of divisors a positive integer $n$ may have, which satisfies the following conditions:
1. $24 \mid n+1$;
2. The sum of the squares of all divisors of $n$ is divisible by $48$ ($1$ and $n$ are included).
To find the smallest possible number of divisors a positive integer \( n \) may have, which satisfies the following conditions:
1. \( 24 \mid n+1 \)
2. The sum of the squares of all divisors of \( n \) is divisible by \( 48 \)

we need to analyze the given conditions and derive the necessary properties of \( n \).

### Step 1: Condition Analysis
1. **Condition 1: \( 24 \mid n+1 \)**
   This implies that \( n+1 \) is a multiple of 24. Therefore, \( n \equiv -1 \pmod{24} \).

2. **Condition 2: Sum of squares of divisors**
   Let \( n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k} \) be the prime factorization of \( n \). The sum of the squares of all divisors of \( n \) is given by:
   \[
   \sigma_2(n) = \prod_{i=1}^k \left( 1 + p_i^2 + p_i^4 + \cdots + p_i^{2\alpha_i} \right)
   \]
   We need \( \sigma_2(n) \) to be divisible by 48.

### Step 2: Divisibility by 48
To ensure \( \sigma_2(n) \) is divisible by 48, we need to consider the prime factors of 48, which are \( 2^4 \times 3 \). Therefore, \( \sigma_2(n) \) must be divisible by \( 2^4 = 16 \) and 3.

### Step 3: Divisibility by 16
For \( \sigma_2(n) \) to be divisible by 16, at least one of the terms in the product must be divisible by 16. Since \( p_i^{2m} \equiv 1 \pmod{16} \) for odd primes \( p_i \), we need to consider the powers of 2 in \( n \).

### Step 4: Divisibility by 3
For \( \sigma_2(n) \) to be divisible by 3, at least one of the terms in the product must be divisible by 3. Since \( p_i^{2m} \equiv 1 \pmod{3} \) for primes \( p_i \neq 3 \), we need to consider the powers of 3 in \( n \).

### Step 5: Combining Conditions
To satisfy both conditions, we need to find the smallest \( n \) such that:
1. \( n \equiv -1 \pmod{24} \)
2. \( \sigma_2(n) \) is divisible by 48

### Step 6: Example and Verification
Consider \( n = 47^{47} \):
1. \( 47 \equiv -1 \pmod{24} \), so \( 47^{47} \equiv (-1)^{47} \equiv -1 \pmod{24} \).
2. The sum of the squares of the divisors of \( 47^{47} \) is:
   \[
   \sigma_2(47^{47}) = 1 + 47^2 + 47^4 + \cdots + 47^{94}
   \]
   Since \( 47 \equiv -1 \pmod{48} \), we have:
   \[
   47^2 \equiv 1 \pmod{48}, \quad 47^4 \equiv 1 \pmod{48}, \quad \text{and so on}
   \]
   Therefore, \( \sigma_2(47^{47}) \equiv 47 \equiv -1 \pmod{48} \), which is divisible by 48.

### Step 7: Number of Divisors
The number of divisors of \( n = 47^{47} \) is \( 47 + 1 = 48 \).

Thus, the smallest possible number of divisors \( n \) may have is \( 48 \).

The final answer is \(\boxed{48}\)