Find all pairs of naturals $a,b$ with $a <b$, such that the sum of the naturals greater than $a$ and less than $ b$ equals $1998$.
1. We start with the given condition that the sum of the naturals greater than \(a\) and less than \(b\) equals \(1998\). This can be written as:
   \[
   \sum_{k=a+1}^{b-1} k = 1998
   \]

2. The sum of an arithmetic series from \(a+1\) to \(b-1\) can be calculated using the formula for the sum of an arithmetic series:
   \[
   \sum_{k=a+1}^{b-1} k = \frac{(a+1) + (b-1)}{2} \cdot (b - a - 1)
   \]

3. Substituting the given sum:
   \[
   \frac{(a+1) + (b-1)}{2} \cdot (b - a - 1) = 1998
   \]

4. Simplifying the expression inside the sum:
   \[
   \frac{a + b}{2} \cdot (b - a - 1) = 1998
   \]

5. Multiplying both sides by 2 to clear the fraction:
   \[
   (a + b) \cdot (b - a - 1) = 3996
   \]

6. We need to find pairs \((a, b)\) such that \(a < b\) and the product \((a + b)(b - a - 1) = 3996\). Note that \(a + b\) and \(b - a - 1\) must have different parities (one must be even and the other odd).

7. We factorize \(3996\) to find suitable pairs:
   \[
   3996 = 2^2 \times 3 \times 7 \times 19
   \]

8. The factor pairs of \(3996\) are:
   \[
   3996 \times 1, \quad 1332 \times 3, \quad 999 \times 4, \quad 444 \times 9, \quad 333 \times 12, \quad 148 \times 27, \quad 111 \times 36, \quad 108 \times 37
   \]

9. We need to check each pair to ensure \(a\) and \(b\) are natural numbers and \(a < b\):
   - For \((3996, 1)\):
     \[
     a + b = 3996, \quad b - a - 1 = 1 \implies b - a = 2 \implies a = 1997, \quad b = 1999
     \]
   - For \((1332, 3)\):
     \[
     a + b = 1332, \quad b - a - 1 = 3 \implies b - a = 4 \implies a = 664, \quad b = 668
     \]
   - For \((999, 4)\):
     \[
     a + b = 999, \quad b - a - 1 = 4 \implies b - a = 5 \implies a = 497, \quad b = 502
     \]
   - For \((444, 9)\):
     \[
     a + b = 444, \quad b - a - 1 = 9 \implies b - a = 10 \implies a = 217, \quad b = 227
     \]
   - For \((333, 12)\):
     \[
     a + b = 333, \quad b - a - 1 = 12 \implies b - a = 13 \implies a = 160, \quad b = 173
     \]
   - For \((148, 27)\):
     \[
     a + b = 148, \quad b - a - 1 = 27 \implies b - a = 28 \implies a = 60, \quad b = 88
     \]
   - For \((111, 36)\):
     \[
     a + b = 111, \quad b - a - 1 = 36 \implies b - a = 37 \implies a = 37, \quad b = 74
     \]
   - For \((108, 37)\):
     \[
     a + b = 108, \quad b - a - 1 = 37 \implies b - a = 38 \implies a = 35, \quad b = 73
     \]

10. Therefore, the pairs \((a, b)\) that satisfy the given condition are:
    \[
    (1997, 1999), (664, 668), (497, 502), (217, 227), (160, 173), (60, 88), (37, 74), (35, 73)
    \]

The final answer is \( \boxed{ \{(1997, 1999), (664, 668), (497, 502), (217, 227), (160, 173), (60, 88), (37, 74), (35, 73)\} } \)