Decide whether there exists a function $f : Z \rightarrow Z$ such that for each $k =0,1, ...,1996$ and for any integer $m$ the equation $f (x)+kx = m$ has at least one integral solution $x$.
To determine whether there exists a function \( f : \mathbb{Z} \rightarrow \mathbb{Z} \) such that for each \( k = 0, 1, \ldots, 1996 \) and for any integer \( m \), the equation \( f(x) + kx = m \) has at least one integral solution \( x \), we proceed as follows:

1. **Rewriting the Equation:**
   The given equation is \( f(x) + kx = m \). We can rewrite this as:
   \[
   f(x) = m - kx
   \]
   Our goal is to find a function \( f \) such that for any integer \( m \) and for each \( k \) in the range \( 0, 1, \ldots, 1996 \), there exists an integer \( x \) satisfying the equation.

2. **Form of \( x \):**
   Consider \( x \) in the form \( x = 1997n + t \), where \( n \) is an integer and \( t \) is an integer in the range \( 0, 1, \ldots, 1996 \). This form ensures that every integer \( x \) can be represented uniquely.

3. **Defining the Function \( f \):**
   We define the function \( f \) as follows:
   \[
   f(1997n + t) = m - k(1997n + t)
   \]
   This definition ensures that for any integer \( m \) and for each \( k \), there exists an integer \( x = 1997n + t \) such that \( f(x) + kx = m \).

4. **Verification:**
   To verify, substitute \( x = 1997n + t \) into the equation:
   \[
   f(1997n + t) + k(1997n + t) = m
   \]
   Using the definition of \( f \):
   \[
   (m - k(1997n + t)) + k(1997n + t) = m
   \]
   Simplifying, we get:
   \[
   m - k(1997n + t) + k(1997n + t) = m
   \]
   \[
   m = m
   \]
   This holds true for any integer \( m \), confirming that the function \( f \) defined in this manner satisfies the given condition.

The final answer is \( \boxed{ f(1997n + t) = m - k(1997n + t) } \)