Given are $N$ points on the plane such that no three of them are collinear,which are coloured red,green and black.We consider all the segments between these points and give to each segment a [i]"value"[/i] according to the following conditions:
[b]i.[/b]If at least one of the endpoints of a segment is black then the segment's [i]"value"[/i] is $0$.
[b]ii.[/b]If the endpoints of the segment have the same colour,re or green,then the segment's [i]"value"[/i] is $1$.
[b]iii.[/b]If the endpoints of the segment have different colours but none of them is black,then the segment's [i]"value"[/i] is $-1$.

Determine the minimum possible sum of the [i]"values"[/i] of the segments.
1. **Define Variables:**
   Let \( r \) be the number of red points, \( g \) be the number of green points, and \( b \) be the number of black points. Thus, the total number of points is \( N = r + g + b \).

2. **Calculate Segments with Black Points:**
   Segments with at least one black endpoint have a value of 0. The total number of segments is \( \binom{N}{2} \). The number of segments with no black endpoints is \( \binom{r+g}{2} \). Therefore, the number of segments with at least one black endpoint is:
   \[
   \binom{N}{2} - \binom{r+g}{2}
   \]
   These segments contribute \( 0 \) to the total sum.

3. **Calculate Segments with Same Color (Red or Green):**
   Segments with both endpoints red or both endpoints green have a value of 1. The number of such segments is:
   \[
   \binom{r}{2} + \binom{g}{2}
   \]
   These segments contribute \( 1 \cdot \left( \binom{r}{2} + \binom{g}{2} \right) \) to the total sum.

4. **Calculate Segments with Different Colors (Red and Green):**
   Segments with one endpoint red and one endpoint green have a value of -1. The number of such segments is:
   \[
   rg
   \]
   These segments contribute \( -1 \cdot rg \) to the total sum.

5. **Total Sum of Values:**
   Combining all contributions, the total sum of values is:
   \[
   0 \cdot \left( \binom{N}{2} - \binom{r+g}{2} \right) + 1 \cdot \left( \binom{r}{2} + \binom{g}{2} \right) - rg
   \]
   Simplifying, we get:
   \[
   \binom{r}{2} + \binom{g}{2} - rg = \frac{r(r-1)}{2} + \frac{g(g-1)}{2} - rg
   \]
   \[
   = \frac{r^2 - r + g^2 - g - 2rg}{2}
   \]

6. **Minimize the Expression:**
   We need to minimize:
   \[
   \frac{r^2 - r + g^2 - g - 2rg}{2}
   \]
   This can be rewritten as:
   \[
   \frac{(r-g)^2 - (r+g)}{2}
   \]
   Since \( r + g = N - b \), the expression becomes:
   \[
   \frac{(r-g)^2 - (N - b)}{2}
   \]
   To minimize this, we need to minimize \( (r-g)^2 \). The minimum value of \( (r-g)^2 \) is 0 when \( r = g \). If \( N \) is even, we can set \( r = g = \frac{N}{2} \). If \( N \) is odd, we can set \( r = \frac{N-1}{2} \) and \( g = \frac{N+1}{2} \) or vice versa.

7. **Final Calculation:**
   - If \( N \) is even:
     \[
     \frac{(r-g)^2 - (N - b)}{2} = \frac{0 - N}{2} = -\frac{N}{2}
     \]
   - If \( N \) is odd:
     \[
     \frac{(r-g)^2 - (N - b)}{2} = \frac{1 - N}{2}
     \]

The final answer is \( \boxed{ -\frac{N}{2} } \) if \( N \) is even, and \(\frac{1-N}{2}\) if \( N \) is odd.