For all integers $n\ge 2$ with the following property:

[list]
[*] for each pair of positive divisors $k,~\ell <n$, at least one of the numbers $2k-\ell$ and $2\ell-k$ is a (not necessarily positive) divisor of $n$ as well.[/list]
1. **Definition and Initial Observations:**
   - We define a positive integer \( n \) as a *pretty* number if for each pair of positive divisors \( k, \ell < n \), at least one of the numbers \( 2k - \ell \) and \( 2\ell - k \) is a (not necessarily positive) divisor of \( n \).
   - If \( k = \ell \), then \( 2k - \ell = 2\ell - k = k \), which is a divisor of \( n \).

2. **Case 1: \( n \) is a Prime Number:**
   - If \( n \) is a prime, the only proper divisor of \( n \) is 1. Thus, \( k = \ell = 1 \) is the only possibility.
   - For \( k = \ell = 1 \), \( 2k - \ell = 2 \cdot 1 - 1 = 1 \), which is a divisor of \( n \).
   - Hence, all prime numbers are pretty numbers.

3. **Case 2: \( n \) is an Even Number Greater than 2:**
   - Let \( n \) be even and \( n > 2 \). Consider \( k = \frac{n}{2} \) and \( \ell = 1 \), which are proper divisors of \( n \).
   - We need \( 2k - \ell = n - 1 \) or \( 2\ell - k = 2 - \frac{n}{2} \) to be a divisor of \( n \).
   - This implies \( \frac{n}{n-1} \) or \( \frac{2n}{n-4} \) must be an integer.
   - Therefore, \( n-1 \mid 1 \) or \( n-4 \mid 8 \), giving us \( n \in \{6, 8, 12\} \).
   - Checking these values:
     - \( n = 6 \): \( k = 3 \), \( \ell = 1 \) gives \( 2k - \ell = 5 \) and \( 2\ell - k = -1 \), neither of which are divisors of 6. However, for other pairs, the conditions hold, so 6 is a pretty number.
     - \( n = 8 \): \( k = 2 \), \( \ell = 1 \) gives \( 2k - \ell = 3 \) and \( 2\ell - k = 0 \), neither of which are divisors of 8. Thus, 8 is not a pretty number.
     - \( n = 12 \): \( k = 6 \), \( \ell = 3 \) gives \( 2k - \ell = 9 \) and \( 2\ell - k = 0 \), neither of which are divisors of 12. Thus, 12 is not a pretty number.
   - Hence, the only even pretty number greater than 2 is 6.

4. **Case 3: \( n \) is a Composite Odd Number:**
   - Let \( n \) be a composite odd integer. Let \( p \) be the smallest prime divisor of \( n \).
   - Consider \( k = \frac{n}{p} \) and \( \ell = 1 \), which are proper divisors of \( n \).
   - We need \( 2k - \ell = 2\frac{n}{p} - 1 \) or \( 2\ell - k = 2 - \frac{n}{p} \) to be a divisor of \( n \).
   - This implies \( \frac{n}{2\frac{n}{p} - 1} \) or \( \frac{n}{\frac{n}{p} - 2} \) must be an integer.

5. **Subcase 1: \( \frac{n}{2\frac{n}{p} - 1} \in \mathbb{Z} \):**
   - Let \( d_1 = \gcd(n, 2\frac{n}{p} - 1) \).
   - Assume \( q \neq p \) is a prime divisor of \( d_1 \). Then \( q \mid n \) and \( q \mid 2\frac{n}{p} - 1 \).
   - Since \( p \) and \( q \) are distinct primes, \( q \mid \frac{n}{p} \), implying \( q \mid 1 \), a contradiction.
   - Thus, \( p \) is the only prime divisor of \( n \).
   - Assume \( p^2 \mid d_1 \). Then \( p^2 \mid n \) and \( p \mid 2\frac{n}{p} - 1 \), implying \( p \mid 1 \), a contradiction.
   - Hence, \( d_1 \in \{1, p\} \).
   - The fact that \( \frac{n}{2\frac{n}{p} - 1} \in \mathbb{Z} \) and \( d_1 = \gcd(n, 2\frac{n}{p} - 1) \) implies \( 2\frac{n}{p} - 1 = d_1 \).
   - Therefore, \( 2\frac{n}{p} - 1 = 1 \) or \( 2\frac{n}{p} - 1 = p \), i.e., \( n = p \) (impossible since \( n \) is not a prime) or \( n = \frac{p(p+1)}{2} \).
   - But \( \frac{p+1}{2} \) must have a prime divisor \( \leq \frac{p+1}{2} \), implying \( \frac{p+1}{2} \geq p \), a contradiction.
   - Thus, \( \frac{n}{2\frac{n}{p} - 1} \not\in \mathbb{Z} \).

6. **Subcase 2: \( \frac{n}{\frac{n}{p} - 2} \in \mathbb{Z} \):**
   - Let \( d_2 = \gcd(n, \frac{n}{p} - 2) \).
   - Assume \( q \neq p \) is a prime divisor of \( d_2 \). Then \( q \mid n \) and \( q \mid \frac{n}{p} - 2 \).
   - Since \( p \) and \( q \) are distinct primes, \( q \mid \frac{n}{p} \), implying \( q \mid 2 \), a contradiction.
   - Assume \( p^2 \mid d_2 \). Then \( p^2 \mid n \) and \( p \mid \frac{n}{p} - 2 \), implying \( p \mid 2 \), a contradiction.
   - Thus, \( d_2 \in \{1, p\} \).
   - The fact that \( d_2 = \frac{n}{p} - 2 \) results in \( n \in \{3p, p(p+2)\} \).

7. **Checking the Possibilities:**
   - **(I) \( n = 3p \):**
     - \( k = p \) and \( \ell = 1 \) are proper divisors of \( n \).
     - We need \( 2k - \ell = 2p - 1 \) or \( 2\ell - k = 2 - p \) to be a divisor of \( n = 3p \).
     - This implies \( 2p - 1 \mid 3p \) or \( p - 2 \mid 3p \), implying \( 2p - 1 \mid 3 \) or \( p - 2 \mid 6 \).
     - Consequently, \( p \in \{1, 2, 3, 4, 5, 8\} \), yielding \( p = 3 \) or \( p = 5 \) because \( p \) is an odd prime.
     - Both \( n = 3 \cdot 3 = 9 \) and \( n = 3 \cdot 5 = 15 \) are pretty numbers.
   - **(II) \( n = p(p+2) \):**
     - \( k = p+2 \) and \( \ell = p \) are proper divisors of \( n \).
     - We need \( 2k - \ell = 2(p+2) - p = p+4 \) or \( 2\ell - k = 2p - (p+2) = p-2 \) to be a divisor of \( n = p(p+2) \).
     - This implies \( p+4 \mid p(p+2) \) or \( p-2 \mid p(p+2) \).
     - Hence, \( p+4 \mid 8 \) or \( p-2 \mid 8 \), yielding \( p = 3 \) and \( n = 3 \cdot 5 = 15 \), which is a pretty number.

**Conclusion:**
The only pretty numbers are primes and the composite numbers 6, 9, and 15.

The final answer is \( \boxed{\text{The only pretty numbers are primes and the composite numbers 6, 9, and 15.}} \)