There are $15$ points on the plane, coloured blue, red and green. It is known that the sum of all pairwise distances between red and blue points equals $51$, between red and green points equals $39$, between blue and green points equals $1$. What are the possible amounts of red, blue and green points?
Given the problem, we need to find the possible amounts of red, blue, and green points such that the sum of all pairwise distances between red and blue points equals $51$, between red and green points equals $39$, and between blue and green points equals $1$. Let $r$, $b$, and $g$ be the number of red, blue, and green points respectively. We know that:

\[ r + b + g = 15 \]

We are given the following conditions:
1. The sum of all pairwise distances between red and blue points is $51$.
2. The sum of all pairwise distances between red and green points is $39$.
3. The sum of all pairwise distances between blue and green points is $1$.

Let's denote the distances between red and blue points as $d_{RB}$, between red and green points as $d_{RG}$, and between blue and green points as $d_{BG}$. We can write the following equations based on the given conditions:

\[ r \cdot b \cdot d_{RB} = 51 \]
\[ r \cdot g \cdot d_{RG} = 39 \]
\[ b \cdot g \cdot d_{BG} = 1 \]

We need to find the natural number solutions for $r$, $b$, and $g$ that satisfy these equations along with the constraint $r + b + g = 15$.

1. **Analyzing the equations:**

   From the third equation:
   \[ b \cdot g \cdot d_{BG} = 1 \]
   Since $d_{BG}$ is a distance, it must be positive. Therefore, $b \cdot g$ must be a divisor of $1$. The only natural number pairs $(b, g)$ that satisfy this are $(1, 1)$.

2. **Substituting $b = 1$ and $g = 1$:**

   \[ r + 1 + 1 = 15 \]
   \[ r = 13 \]

   Now, we need to check if these values satisfy the other two equations:
   \[ 13 \cdot 1 \cdot d_{RB} = 51 \]
   \[ d_{RB} = \frac{51}{13} \]

   \[ 13 \cdot 1 \cdot d_{RG} = 39 \]
   \[ d_{RG} = \frac{39}{13} \]

   Both $d_{RB}$ and $d_{RG}$ are positive, so the solution $(r, b, g) = (13, 1, 1)$ is valid.

3. **Exploring other possible values:**

   Let's consider other possible values for $b$ and $g$ that satisfy the constraint $r + b + g = 15$.

   If $b = 4$ and $g = 3$:
   \[ r + 4 + 3 = 15 \]
   \[ r = 8 \]

   Now, we need to check if these values satisfy the other two equations:
   \[ 8 \cdot 4 \cdot d_{RB} = 51 \]
   \[ d_{RB} = \frac{51}{32} \]

   \[ 8 \cdot 3 \cdot d_{RG} = 39 \]
   \[ d_{RG} = \frac{39}{24} \]

   \[ 4 \cdot 3 \cdot d_{BG} = 1 \]
   \[ d_{BG} = \frac{1}{12} \]

   All distances are positive, so the solution $(r, b, g) = (8, 4, 3)$ is also valid.

Conclusion:
The possible amounts of red, blue, and green points are $(13, 1, 1)$ and $(8, 4, 3)$.

The final answer is $\boxed{(13, 1, 1), (8, 4, 3)}$