Let f be a function that satisfies the following conditions:

$(i)$ If $x > y$ and $f(y) - y \geq v \geq f(x) - x$, then $f(z) = v + z$, for some number $z$ between $x$ and $y$.
$(ii)$ The equation $f(x) = 0$ has at least one solution, and among the solutions of this equation, there is one that is not smaller than all the other solutions;
$(iii)$ $f(0) = 1$.
$(iv)$ $f(1987) \leq 1988$.
$(v)$ $f(x)f(y) = f(xf(y) + yf(x) - xy)$.

Find $f(1987)$.

[i]Proposed by Australia.[/i]
1. **Verify the given function \( f(x) = x + 1 \) against all conditions:**

   - **Condition (i):** If \( x > y \) and \( f(y) - y \geq v \geq f(x) - x \), then \( f(z) = v + z \) for some \( z \) between \( x \) and \( y \).

     For \( f(x) = x + 1 \):
     \[
     f(y) - y = (y + 1) - y = 1
     \]
     \[
     f(x) - x = (x + 1) - x = 1
     \]
     Since \( 1 \geq v \geq 1 \), we have \( v = 1 \). Therefore, \( f(z) = 1 + z \), which is consistent with \( f(z) = z + 1 \).

   - **Condition (ii):** The equation \( f(x) = 0 \) has at least one solution, and among the solutions of this equation, there is one that is not smaller than all the other solutions.

     For \( f(x) = x + 1 \):
     \[
     x + 1 = 0 \implies x = -1
     \]
     The solution \( x = -1 \) is unique and satisfies the condition.

   - **Condition (iii):** \( f(0) = 1 \).

     For \( f(x) = x + 1 \):
     \[
     f(0) = 0 + 1 = 1
     \]
     This condition is satisfied.

   - **Condition (iv):** \( f(1987) \leq 1988 \).

     For \( f(x) = x + 1 \):
     \[
     f(1987) = 1987 + 1 = 1988
     \]
     This condition is satisfied.

   - **Condition (v):** \( f(x)f(y) = f(xf(y) + yf(x) - xy) \).

     For \( f(x) = x + 1 \):
     \[
     f(x)f(y) = (x + 1)(y + 1) = xy + x + y + 1
     \]
     \[
     f(xf(y) + yf(x) - xy) = f(x(y + 1) + y(x + 1) - xy) = f(xy + x + y + yx - xy) = f(x + y + y) = f(x + y + y) = x + y + y + 1
     \]
     Both sides are equal, so this condition is satisfied.

2. **Conclusion:**

   Since \( f(x) = x + 1 \) satisfies all the given conditions, we can conclude that \( f(1987) = 1987 + 1 = 1988 \).

The final answer is \( \boxed{1988} \).