Find all functions $ f: Z \rightarrow R$ that verify the folowing two conditions:
(i) for each pair of integers $ (m,n)$ with $ m<n$ one has $ f(m)<f(n)$;
(ii) for each pair of integers $ (m,n)$ there exists an integer $ k$ such that $ f(m)\minus{}f(n)\equal{}f(k)$.
1. **Condition (i):** For each pair of integers \( (m, n) \) with \( m < n \), we have \( f(m) < f(n) \). This implies that \( f \) is a strictly increasing function.

2. **Condition (ii):** For each pair of integers \( (m, n) \), there exists an integer \( k \) such that \( f(m) - f(n) = f(k) \). This implies that the difference between any two function values is itself a function value.

3. **Scaling Argument:** If \( f \) satisfies the conditions, then \( \frac{f(x)}{a} \) also satisfies the conditions for any positive number \( a \). Therefore, we can assume without loss of generality that there exists an integer \( n \) such that \( f(n) = 1 \).

4. **Set of Values:** Let \( M \) be the set of values of the function \( f \). From condition (ii), we have \( f(m) - 1 \in M \). Additionally, for every \( k \), there exists an integer \( k' \) such that \( f(k') = -f(k) \). This follows from changing \( m \) to \( n \) in condition (ii).

5. **Induction Argument:** By induction, we can show that \( \mathbb{Z} \subseteq M \). This means that every integer is a value of the function \( f \).

6. **Symmetry in Values:** From condition (ii), we have \( t \in M \Leftrightarrow -t \in M \). This implies that the set \( M \) is symmetric about the origin.

7. **Arithmetic Sequence:** Consider the set of values \( \{f(m), f(m+1), \ldots, f(m+k)\} \). We will prove that this set must form an arithmetic sequence. 

   - We have \( f(m+k) - f(m) > f(m+k) - f(m+1) > \ldots > f(m+k) - f(m+k-1) \).
   - This implies \( f(m+k) > f(m+k-1) > \ldots > f(m) \).

   Since the differences between consecutive terms are equal, the sequence must be arithmetic. 

8. **Form of the Function:** From the above, we conclude that \( f(n) = an + b \) for some constants \( a \) and \( b \). 

9. **Determining Constants:** Since there exists \( k \) such that \( ak + b = 0 \), we have \( b = -ak \) for some integer \( k \). Therefore, \( f(n) = a(n + t) \) where \( t \) is an integer and \( a \) is a positive real number.

The final answer is \( \boxed{ f(n) = a(n + t) } \) where \( a \) is a positive real number and \( t \) is an integer.