There are 2008 congruent circles on a plane such that no two are tangent to each other and each circle intersects at least three other circles. Let $ N$ be the total number of intersection points of these circles. Determine the smallest possible values of $ N$.
1. **Understanding the Problem:**
   We have 2008 congruent circles on a plane, where no two circles are tangent to each other, and each circle intersects with at least three other circles. We need to determine the smallest possible value of \( N \), the total number of intersection points of these circles.

2. **Intersection Points:**
   Each intersection point is shared by exactly two circles. If a point belongs to \( n \) circles, then each of these circles gets a score of \( \frac{1}{n} \) from this point. We need to ensure that the sum of scores on each circle is at least 1.

3. **Score Calculation:**
   Let \( A_1, A_2, \ldots, A_t \) be points on the same circle, and each point belongs to at least two circles. The score of each circle from these points is \( \frac{1}{t} \times t = 1 \).

4. **Minimum Number of Intersection Points:**
   Since each circle intersects with at least three other circles, each circle must have at least three intersection points. Therefore, the minimum number of intersection points per circle is 3.

5. **Total Number of Intersection Points:**
   Since there are 2008 circles, and each circle intersects with at least three other circles, the total number of intersection points \( N \) must be at least \( 2008 \times 3 / 2 \) (since each intersection point is counted twice, once for each circle).

   \[
   N \geq \frac{2008 \times 3}{2} = 3012
   \]

6. **Verification:**
   To verify, consider the configuration where each circle intersects exactly three other circles. In this case, each intersection point is shared by exactly two circles, and the total number of intersection points is minimized.

\[
N = 3012
\]

The final answer is \( \boxed{3012} \).