The real-valued function $f$ is defined for positive integers, and the positive integer $a$ satisfies
$f(a) = f(1995), f(a+1) = f(1996), f(a+2) = f(1997), f(n + a) = \frac{f(n) - 1}{f(n) + 1}$ for all positive integers $n$.
(i) Show that $f(n+ 4a) = f(n)$ for all positive integers $n$.
(ii) Determine the smallest possible $a$.
### Part (i)
1. We start with the given functional equation:
   \[
   f(n + a) = \frac{f(n) - 1}{f(n) + 1}
   \]
   for all positive integers \( n \).

2. To show that \( f(n + 4a) = f(n) \), we will iterate the functional equation multiple times.

3. First, apply the functional equation to \( n + a \):
   \[
   f(n + 2a) = \frac{f(n + a) - 1}{f(n + a) + 1}
   \]
   Substitute \( f(n + a) = \frac{f(n) - 1}{f(n) + 1} \):
   \[
   f(n + 2a) = \frac{\frac{f(n) - 1}{f(n) + 1} - 1}{\frac{f(n) - 1}{f(n) + 1} + 1}
   \]
   Simplify the expression:
   \[
   f(n + 2a) = \frac{\frac{f(n) - 1 - (f(n) + 1)}{f(n) + 1}}{\frac{f(n) - 1 + (f(n) + 1)}{f(n) + 1}} = \frac{\frac{f(n) - 1 - f(n) - 1}{f(n) + 1}}{\frac{f(n) - 1 + f(n) + 1}{f(n) + 1}} = \frac{\frac{-2}{f(n) + 1}}{\frac{2f(n)}{f(n) + 1}} = -\frac{1}{f(n)}
   \]
   Thus, we have:
   \[
   f(n + 2a) = -\frac{1}{f(n)}
   \]

4. Next, apply the functional equation to \( n + 2a \):
   \[
   f(n + 3a) = \frac{f(n + 2a) - 1}{f(n + 2a) + 1}
   \]
   Substitute \( f(n + 2a) = -\frac{1}{f(n)} \):
   \[
   f(n + 3a) = \frac{-\frac{1}{f(n)} - 1}{-\frac{1}{f(n)} + 1} = \frac{-\frac{1 + f(n)}{f(n)}}{-\frac{1 - f(n)}{f(n)}} = \frac{-(1 + f(n))}{-(1 - f(n))} = \frac{1 + f(n)}{1 - f(n)}
   \]

5. Finally, apply the functional equation to \( n + 3a \):
   \[
   f(n + 4a) = \frac{f(n + 3a) - 1}{f(n + 3a) + 1}
   \]
   Substitute \( f(n + 3a) = \frac{1 + f(n)}{1 - f(n)} \):
   \[
   f(n + 4a) = \frac{\frac{1 + f(n)}{1 - f(n)} - 1}{\frac{1 + f(n)}{1 - f(n)} + 1} = \frac{\frac{1 + f(n) - (1 - f(n))}{1 - f(n)}}{\frac{1 + f(n) + (1 - f(n))}{1 - f(n)}} = \frac{\frac{2f(n)}{1 - f(n)}}{\frac{2}{1 - f(n)}} = f(n)
   \]

6. Therefore, we have shown that:
   \[
   f(n + 4a) = f(n)
   \]
   for all positive integers \( n \).

### Part (ii)
1. We need to determine the smallest possible \( a \) such that the function \( f \) satisfies the given conditions.

2. From part (i), we know that \( f \) is periodic with period \( 4a \). We need to check the smallest \( a \) that satisfies the conditions.

3. Suppose \( a = 1 \):
   \[
   f(n + 1) = \frac{f(n) - 1}{f(n) + 1}
   \]
   Then \( f \) would be periodic with period 4. However, \( f(1) = f(1995) \) and \( 1995 \equiv 3 \pmod{4} \). This implies:
   \[
   f(1) = f(3) = -\frac{1}{f(1)}
   \]
   which leads to:
   \[
   f(1)^2 = -1
   \]
   This is impossible for real-valued functions. Hence, \( a = 1 \) is not possible.

4. Suppose \( a = 2 \):
   \[
   f(n + 2) = \frac{f(n) - 1}{f(n) + 1}
   \]
   Then \( f \) would be periodic with period 8. However, \( f(2) = f(1996) \) and \( 1996 \equiv 4 \pmod{8} \). This implies:
   \[
   f(2) = f(4) = \frac{f(2) - 1}{f(2) + 1}
   \]
   which leads to:
   \[
   f(2)^2 = -1
   \]
   This is also impossible for real-valued functions. Hence, \( a = 2 \) is not possible.

5. Suppose \( a = 3 \):
   \[
   f(n + 3) = \frac{f(n) - 1}{f(n) + 1}
   \]
   Then \( f \) would be periodic with period 12. We can choose \( f(1) = f(2) = f(3) = 2 \) (or any other number) and compute \( f \) depending on \( n \mod 12 \) by the recursion. This does not lead to any contradictions.

6. Therefore, the smallest possible \( a \) is:
   \[
   \boxed{3}
   \]