Find all integers $k$ for which,  there is a function $f: N \to Z$ that satisfies:
(i) $f(1995) = 1996$
(ii) $f(xy) = f(x) + f(y) + kf(m_{xy})$ for all natural numbers $x, y$,where$ m_{xy}$ denotes the greatest common divisor of the numbers $x, y$.

Clarification: $N = \{1,2,3,...\}$ and $Z = \{...-2,-1,0,1,2,...\}$ .
1. **Given Conditions:**
   - \( f: \mathbb{N} \to \mathbb{Z} \)
   - \( f(1995) = 1996 \)
   - \( f(xy) = f(x) + f(y) + k f(\gcd(x, y)) \) for all natural numbers \( x, y \)

2. **Analyzing the Functional Equation:**
   - Let \( n = p^\alpha \), where \( p \) is a prime and \( \alpha \geq 1 \).
   - Using induction on \( \alpha \), we get:
     \[
     f(p^\alpha) = (\alpha k - k + \alpha) f(p)
     \]

3. **Substituting Specific Values:**
   - Let \( x = p^2 \) and \( y = p^3 \) for some prime \( p \) such that \( f(p) \neq 0 \).
   - Substituting into the functional equation:
     \[
     f(p^5) = f(p^2 \cdot p^3) = f(p^2) + f(p^3) + k f(\gcd(p^2, p^3))
     \]
   - Since \( \gcd(p^2, p^3) = p^2 \), we have:
     \[
     f(p^5) = f(p^2) + f(p^3) + k f(p^2)
     \]
   - Using the earlier result \( f(p^\alpha) = (\alpha k - k + \alpha) f(p) \):
     \[
     (5k - k + 5) f(p) = (2k - k + 2) f(p) + (3k - k + 3) f(p) + k (2k - k + 2) f(p)
     \]
   - Simplifying:
     \[
     (4k + 5) f(p) = (k + 2) f(p) + (2k + 3) f(p) + k (k + 2) f(p)
     \]
     \[
     4k + 5 = k + 2 + 2k + 3 + k^2 + 2k
     \]
     \[
     4k + 5 = k^2 + 5k + 5
     \]
     \[
     0 = k^2 + k
     \]
     \[
     k(k + 1) = 0
     \]
   - Therefore, \( k = -1 \) or \( k = 0 \).

4. **Case \( k = -1 \):**
   - Consider the constant function \( f(x) \equiv 1996 \).
   - This function satisfies:
     \[
     f(xy) = 1996 = 1996 + 1996 - 1996 = f(x) + f(y) - f(\gcd(x, y))
     \]
   - Hence, \( f(x) = 1996 \) is a valid solution.

5. **Case \( k = 0 \):**
   - The functional equation simplifies to:
     \[
     f(xy) = f(x) + f(y)
     \]
   - This is a Cauchy-type functional equation for natural numbers.
   - If \( n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_t^{\alpha_t} \), then:
     \[
     f(n) = f(p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_t^{\alpha_t}) = \sum \alpha_i f(p_i)
     \]
   - Define \( f \) for primes and \( x = 1 \):
     \[
     f(x) = \begin{cases} 
     0 & \text{if } x = 1 \\
     1981 & \text{if } x = 19 \\
     1 & \text{if } x \in P \setminus \{19\}
     \end{cases}
     \]
   - Verify:
     \[
     1995 = 3 \cdot 5 \cdot 7 \cdot 19
     \]
     \[
     f(1995) = f(3) + f(5) + f(7) + f(19) = 1 + 1 + 1 + 1981 = 1984
     \]
   - Adjust \( f \) to satisfy \( f(1995) = 1996 \):
     \[
     f(19) = 1993
     \]
   - Thus, \( f(x) \) satisfies the conditions.

The final answer is \( \boxed{ k = -1 } \) or \( k = 0 \).