From the three rules for the determination of the algebraic value of the line segments we have the following table:

Let now that we have $\kappa$ red, $\pi$ green and $\mu$ black points. Then it is clear that $\kappa + \pi + \mu = \nu$.

The $\kappa$ red points determine $\binom{\kappa}{2}$ line segments having their ends red and so they have algebraic values 1. The $\pi$ green points determine $\binom{\pi}{2}$ line segments with both ends green and therefore with algebraic values 1. The number of the line segments having ends with different colors, red or green, and so with algebraic value -1, are $\kappa \cdot \pi$. All the other line segments have algebraic value 0, because they have at least one of their ends black.

The sum of the algebraic values of all existing line segments is:
$$\begin{aligned} \Sigma &= \binom{\kappa}{2} + \binom{\pi}{2} - \kappa\pi = \frac{\kappa!}{(\kappa-2)!2!} + \frac{\pi!}{(\pi-2)!2!} - \kappa\pi = \\ &= \frac{\kappa(\kappa-1)}{2} + \frac{\pi(\pi-1)}{2} - \frac{2\kappa\pi}{2} = \frac{\kappa^2 - \kappa}{2} + \frac{\pi^2 - \pi}{2} - \frac{2\kappa\pi}{2} = \\ &= \frac{(\kappa - \pi)^2}{2} - \frac{\kappa + \pi}{2} \quad (\text{because } \kappa + \pi + \mu = \nu) \\ &= \frac{(\kappa - \pi)^2}{2} - \frac{\nu - \mu}{2} = \frac{(\kappa - \pi)^2}{2} + \frac{\mu}{2} - \frac{\nu}{2}. \end{aligned}$$

From the last expression of $\Sigma$ we conclude that:
$$\Sigma \geq -\frac{\nu}{2} \quad (1)$$

If $\nu$ is even (let $\nu = 2\rho$), then relation (1) becomes: $\Sigma \geq -\rho$.
Equality in the last relation holds, if and only if $\kappa = \pi = \frac{\nu}{2} = \rho$ and $\mu = 0$.
For example, for $\nu = 4$, we have the following result:

Least total sum: $\Sigma = -\frac{\nu}{2} = -\frac{4}{2} = -2$

If $\nu$ is odd ($\nu = 2\rho + 1$), then relation (1) $\gamma\dot{\iota}\nu\epsilon\tau\alpha\iota$:
$$\Sigma \geq -\frac{2\rho + 1}{2} = -\rho - \frac{1}{2}$$
Since $\Sigma$ is integer we conclude that: $\Sigma \geq -\rho = \frac{-\nu + 1}{2}$.

We check now when equality holds in the last relation. We observe that the case “$\nu$ odd ($\nu = 2\rho + 1$)” comes from the case “$\nu$ even ($\nu = 2\rho$)” by adding one more point. The point we add in the case “$\nu$ even ($\nu = 2\rho$)” can be blank, red or green, and so we have the following cases:

### Case 1
Let the point we add is blank. Then the new produced line segments have algebraic values 0 and the equality in this case holds when: $\kappa = \pi = \frac{\nu - 1}{2}$ and $\mu = 1$.
The sum of all algebraic values is: $\Sigma = -\rho = \frac{-\nu + 1}{2}$.

Least total sum: $\Sigma = \frac{-\nu + 1}{2} = \frac{-5 + 1}{2} = -2$

### Case 2
Let the point we add is red. We had red and green points: $\kappa = \pi = \frac{\nu-1}{2}$. Now with the new point we can create $\frac{\nu-1}{2}$ line segments having algebraic value 1 and $\frac{\nu-1}{2}$ line segments having algebraic value -1. Equality in this case holds when
$$\kappa = \frac{\nu + 1}{2}, \pi = \frac{\nu - 1}{2} \text{ and } \mu = 0.$$
The sum of all algebraic values remains: $\Sigma = -\rho = \frac{-\nu + 1}{2}$.

### Case 3
If the point we add is green, in a similar way we conclude that the equality holds when $\kappa = \frac{\nu - 1}{2}$, $\pi = \frac{\nu + 1}{2}$ and $\mu = 0$. Again: $\Sigma = -\rho = \frac{-\nu + 1}{2}$.

From all the above we conclude that the least possible value of $\Sigma$ is $-\lfloor \frac{\nu}{2} \rfloor$.