Let $f$ be a function defined on the set of positive integers , and with values in the same set, which satisfies:
$\bullet$ $f (n + f (n)) = 1$ for all $n\ge 1$.
$\bullet$ $f (1998) = 2$
Find the lowest possible value of the sum $f (1) + f (2) +... + f (1999)$, and find the formula of $f$ for which this minimum is satisfied,
1. **Understanding the given conditions:**
   - We are given a function \( f \) defined on the set of positive integers with values in the same set.
   - The function satisfies \( f(n + f(n)) = 1 \) for all \( n \geq 1 \).
   - We are also given \( f(1998) = 2 \).

2. **Analyzing the function:**
   - From the condition \( f(n + f(n)) = 1 \), we know that for any \( n \), \( n + f(n) \) must be such that \( f(n + f(n)) = 1 \).
   - Given \( f(1998) = 2 \), we have \( 1998 + f(1998) = 2000 \), and thus \( f(2000) = 1 \).

3. **Inductive reasoning:**
   - We need to find the minimum possible value of the sum \( f(1) + f(2) + \cdots + f(1999) \).
   - We start by noting that \( f(1997 + f(1997)) = 1 \). Since \( f(1998) = 2 \), it implies \( f(1997) \geq 2 \).
   - By induction, we can show that \( f(n) \geq 2 \) for all \( n \leq 1998 \).

4. **Establishing a lower bound:**
   - We need to ensure that \( f(n) \geq 1999 - n \) for all \( n \leq 1997 \). If \( f(n) < 1999 - n \), then \( n + f(n) \leq 1998 \), which would contradict \( f(n + f(n)) = 1 \).

5. **Summing the values:**
   - We calculate the sum \( f(1) + f(2) + \cdots + f(1999) \) under the assumption that \( f(n) = 1999 - n \) for \( n \leq 1997 \), \( f(1998) = 2 \), and \( f(1999) = 1 \).
   - The sum is:
     \[
     \sum_{k=1}^{1997} (1999 - k) + f(1998) + f(1999)
     \]
   - This can be simplified as:
     \[
     \sum_{k=1}^{1997} 1999 - \sum_{k=1}^{1997} k + 2 + 1
     \]
   - The first sum is:
     \[
     1999 \times 1997 - \sum_{k=1}^{1997} k
     \]
   - The sum of the first 1997 positive integers is:
     \[
     \sum_{k=1}^{1997} k = \frac{1997 \times 1998}{2}
     \]
   - Therefore, the total sum is:
     \[
     1999 \times 1997 - \frac{1997 \times 1998}{2} + 3
     \]
   - Simplifying further:
     \[
     1999 \times 1997 - 998501 + 3 = 1999 \times 1997 - 998501 + 3 = 1999 \times 1997 - 998498
     \]
   - Calculating \( 1999 \times 1997 \):
     \[
     1999 \times 1997 = 1999 \times (2000 - 3) = 1999 \times 2000 - 1999 \times 3 = 3998000 - 5997 = 3992003
     \]
   - Therefore, the sum is:
     \[
     3992003 - 998498 = 1997003
     \]

6. **Verifying the function:**
   - The function \( f(n) = 1999 - n \) for \( n \leq 1997 \), \( f(1998) = 2 \), and \( f(1999) = 1 \) satisfies the given conditions and achieves the minimum sum.

The final answer is \( \boxed{1997003} \).