In a village $1998$ persons volunteered to clean up, for a fair, a rectangular field with integer sides and perimeter equla to $3996$ feet. For this purpose, the field was divided into $1998$ equal parts. If each part had an integer area, find the length and breadth of the field.
1. **Translate the problem into equations:**
   Given the perimeter of the rectangular field is \(3996\) feet, we can write:
   \[
   2(x + y) = 3996 \implies x + y = 1998
   \]
   Also, the field is divided into \(1998\) equal parts, each with an integer area. Let the area of the field be \(A\). Then:
   \[
   A = xy \quad \text{and} \quad \frac{A}{1998} = k \quad \text{(where \(k\) is an integer)}
   \]
   Thus:
   \[
   xy = 1998k
   \]

2. **Formulate the quadratic equation:**
   We can express \(x\) and \(y\) as the roots of the quadratic equation:
   \[
   t^2 - (x+y)t + xy = 0
   \]
   Substituting \(x + y = 1998\) and \(xy = 1998k\), we get:
   \[
   t^2 - 1998t + 1998k = 0
   \]

3. **Solve the quadratic equation:**
   The solutions for \(t\) are given by:
   \[
   t = \frac{1998 \pm \sqrt{1998^2 - 4 \cdot 1998k}}{2}
   \]
   Simplifying inside the square root:
   \[
   t = \frac{1998 \pm \sqrt{1998^2 - 7992k}}{2}
   \]
   Let \(D = 1998^2 - 7992k\). For \(t\) to be an integer, \(D\) must be a perfect square. Let \(D = w^2\):
   \[
   1998^2 - 7992k = w^2
   \]
   Rearranging:
   \[
   1998^2 - w^2 = 7992k
   \]
   Factoring the left-hand side:
   \[
   (1998 - w)(1998 + w) = 7992k
   \]

4. **Find suitable \(w\):**
   We need to find \(w\) such that \((1998 - w)(1998 + w)\) is divisible by \(7992\). Note that:
   \[
   1998 = 2 \times 999 = 2 \times 3^3 \times 37
   \]
   and:
   \[
   7992 = 4 \times 1998 = 2^3 \times 3^3 \times 37
   \]
   We need \(w\) such that both \(1998 - w\) and \(1998 + w\) are divisible by \(2\), \(3\), and \(37\). Trying \(w = 333\):
   \[
   1998 - 333 = 1665 \quad \text{and} \quad 1998 + 333 = 2331
   \]
   Check divisibility:
   \[
   1665 = 3^2 \times 5 \times 37 \quad \text{and} \quad 2331 = 3 \times 777 = 3 \times 3 \times 259 = 3^2 \times 7 \times 37
   \]
   Both are divisible by \(3\) and \(37\).

5. **Calculate dimensions:**
   Using \(w = 333\), we find:
   \[
   x = \frac{1998 + 333}{2} = 1165.5 \quad \text{and} \quad y = \frac{1998 - 333}{2} = 832.5
   \]
   However, these are not integers. Re-evaluate \(w\) to find integer solutions. Trying \(w = 666\):
   \[
   1998 - 666 = 1332 \quad \text{and} \quad 1998 + 666 = 2664
   \]
   Check:
   \[
   x = 1332 \quad \text{and} \quad y = 666
   \]
   Both are integers and satisfy the conditions.

The final answer is \(\boxed{(1332, 666)}\).