The question has since been edited to clarify which numbered balls can be used. I assumed you could use zero, but I guess you can't. Oh well!
Assuming that there's a ball numbered 0, the minimum value for the cuspid is 20, constructed like so:
6
2 1 9
5 0 3 7 4 20
For this proof, I'll use $b_n$ to represent the ball numbered $n$.
First, notice that the balls in the middle layer cannot be $b_0$, $b_1$, or $b_2$; as $0$, $1$, and $2$ cannot be represented as the sum of three unique positive integers. The smallest number with this property is $3$, represented as $0 + 1 + 2$.
To try and minimize the value of the cuspid, let's pick the next two smallest numbered for our middle layer. This makes our middle layer $b_3$, $b_4$, and $b_5$, with a cuspid of $b_{12}$.
But, notice that $b_3$, $b_4$, and $b_5$ cannot make up the middle layer. To see this, let's figure out what numbered balls balls $3$, $4$, and $5$ themselves can sit on. Namely, let's list the partitions of 3, 4, and 5, which are:
— size three
— composed of no duplicates
The only partition of three which satisfies these conditions is $0 + 1 + 2$. That means, $b_3$ can only be placed on $b_0$, $b_1$, and $b_2$.
The only partition of four which satisfies these conditions is $0 + 1 + 3$. That means, $b_4$ can only be placed on $b_0$, $b_1$, and $b_3$.
The only partitions of five which satisfy these conditions are $0 + 1 + 4$ and $0 + 2 + 3$. That means, $b_5$ can only be placed on [$b_0$, $b_1$, and $b_4$] or on [$b_0$, $b_2$, and $b_3$].
Now, notice that it is impossible to construct this, for a few reasons.
1. $b_4$ can only be placed on $b_0$, $b_1$, and $b_3$. This would mean that there are two $b_3$'s. One in the middle layer, since we've assumed so; one in the bottom layer supporting $b_4$. But, since there is only one of each ball, this is not possible.
2. $b_3$ and $b_4$ both require to be placed on $b_0$ and $b_1$. But, because of the topology of our tetrahedron, there is only one ball supporting any two given balls in the middle layer. Therefore, this is not possible.
Therefore, the middle layer cannot be made up of $b_3$, $b_4$, and $b_5$, and the cuspid must be $b_{n}, n > 12$.
Above, I've outlined two main candidates that indicate that a given configuration is impossible.
1. $b_k$ must be directly supported by $b_m$, but we've assumed that $b_m$ is in the middle layer. Since we cannot have two $b_m$'s, this is impossible.
2. $b_y$ and $b_z$ both must be supported by $b_t$ and $b_u$. Since $b_t$ and $b_u$ cannot both directly support $b_y$ and $b_z$, this is impossible.
For example, $b_4$ and $b_6$ cannot simultaneously exist in the middle layer. Of the valid partitions of six, notice that
— $0 + 1 + 5$ shares a "pair" with $0 + 1 + 3$, $(0, 1)$
— $0 + 2 + 4$ requires to be placed on $b_4$, but cannot.
— $1 + 2 + 3$ shares a "pair" with $0 + 1 + 3$, $(1, 3)$
Now that we have these two rules, we can:
i. assume that the cuspid has some given value ($13$, $14$, $15$, etc...)
ii. work through the possible middle-layer values and prove that is is or it isn't possible.
Assuming a cuspid of $b_{13}$; the possible middle-layers are:
— $6 + 4 + 3$, which is not possible because of the pair $(3, 4)$.
Assuming a cuspid of $b_{14}$; the possible middle-layers are:
— $7 + 4 + 3$, which is not possible because of the pair $(3, 4)$.
— $6 + 5 + 3$, which is not possible because of the pair $(3, 5)$.
Assuming a cuspid of $b_{15}$; the possible middle-layers are:
— $8 + 4 + 3$, which is not possible because of the pair $(3, 4)$.
— $7 + 5 + 3$, which is not possible because of the pair $(3, 5)$.
— $6 + 5 + 4$, which is not possible because of the pair $(4, 6)$.
Assuming a cuspid of $b_{16}$; the possible middle-layers are:
— $9 + 4 + 3$, which is not possible because of the pair $(3, 4)$.
— $8 + 5 + 3$, which is not possible because of the pair $(3, 5)$.
— $7 + 6 + 3$, which is not possible because of the pair $(3, 6)$.
— $7 + 5 + 4$, which is not possible because of the pair $(4, 5)$.
Assuming a cuspid of $b_{17}$; the possible middle-layers are:
— $10 + 4 + 3$, which is not possible because of the pair $(3, 4)$.
— $9 + 5 + 3$, which is not possible because of the pair $(3, 5)$.
— $8 + 6 + 3$, which is not possible because of the pair $(3, 6)$.
— $8 + 5 + 4$, which is not possible because of the pair $(4, 5)$.
— $7 + 6 + 4$, which is not possible because of the pair $(4, 6)$.
Assuming a cuspid of $b_{18}$; the possible middle-layers are:
— $11 + 4 + 3$, which is not possible because of the pair $(3, 4)$.
— $10 + 5 + 3$, which is not possible because of the pair $(3, 5)$.
— $9 + 6 + 3$, which is not possible because of the pair $(3, 6)$.
— $9 + 5 + 4$, which is not possible because of the pair $(4, 5)$.
— $8 + 7 + 3$, which is not possible because of the pair $(3, 7)$.
— $8 + 6 + 4$, which is not possible because of the pair $(4, 6)$.
— $7 + 6 + 5$, which is not possible because of the combination of pairs $(5, 6) $ and $(6, 7)$.
It is not hard to continue this logic to find:
$b_{20}$ = $b_{9}$ + $b_{7}$ + $b_{4}$
$b_{9}$ = $b_{6}$ + $b_{2}$ + $b_{1}$
$b_{7}$ = $b_{0}$ + $b_{2}$ + $b_{5}$
$b_{4}$ = $b_{0}$ + $b_{1}$ + $b_{3}$6
2 1 9
5 0 3 7 4 20
For this proof, I'll use $b_n$ to represent the ball numbered $n$.
First, notice that the balls in the middle layer cannot be $b_0$, $b_1$, or $b_2$; as $0$, $1$, and $2$ cannot be represented as the sum of three unique positive integers. The smallest number with this property is $3$, represented as $0 + 1 + 2$.
To try and minimize the value of the cuspid, let's pick the next two smallest numbered for our middle layer. This makes our middle layer $b_3$, $b_4$, and $b_5$, with a cuspid of $b_{12}$.
But, notice that $b_3$, $b_4$, and $b_5$ cannot make up the middle layer. To see this, let's figure out what numbered balls balls $3$, $4$, and $5$ themselves can sit on. Namely, let's list the partitions of 3, 4, and 5, which are:
— size three
— composed of no duplicates
The only partition of three which satisfies these conditions is $0 + 1 + 2$. That means, $b_3$ can only be placed on $b_0$, $b_1$, and $b_2$.
The only partition of four which satisfies these conditions is $0 + 1 + 3$. That means, $b_4$ can only be placed on $b_0$, $b_1$, and $b_3$.
The only partitions of five which satisfy these conditions are $0 + 1 + 4$ and $0 + 2 + 3$. That means, $b_5$ can only be placed on [$b_0$, $b_1$, and $b_4$] or on [$b_0$, $b_2$, and $b_3$].
Now, notice that it is impossible to construct this, for a few reasons.
1. $b_4$ can only be placed on $b_0$, $b_1$, and $b_3$. This would mean that there are two $b_3$'s. One in the middle layer, since we've assumed so; one in the bottom layer supporting $b_4$. But, since there is only one of each ball, this is not possible.
2. $b_3$ and $b_4$ both require to be placed on $b_0$ and $b_1$. But, because of the topology of our tetrahedron, there is only one ball supporting any two given balls in the middle layer. Therefore, this is not possible.
Therefore, the middle layer cannot be made up of $b_3$, $b_4$, and $b_5$, and the cuspid must be $b_{n}, n > 12$.
Above, I've outlined two main candidates that indicate that a given configuration is impossible.
1. $b_k$ must be directly supported by $b_m$, but we've assumed that $b_m$ is in the middle layer. Since we cannot have two $b_m$'s, this is impossible.
2. $b_y$ and $b_z$ both must be supported by $b_t$ and $b_u$. Since $b_t$ and $b_u$ cannot both directly support $b_y$ and $b_z$, this is impossible.
For example, $b_4$ and $b_6$ cannot simultaneously exist in the middle layer. Of the valid partitions of six, notice that
— $0 + 1 + 5$ shares a "pair" with $0 + 1 + 3$, $(0, 1)$
— $0 + 2 + 4$ requires to be placed on $b_4$, but cannot.
— $1 + 2 + 3$ shares a "pair" with $0 + 1 + 3$, $(1, 3)$
Now that we have these two rules, we can:
i. assume that the cuspid has some given value ($13$, $14$, $15$, etc...)
ii. work through the possible middle-layer values and prove that is is or it isn't possible.
Assuming a cuspid of $b_{13}$; the possible middle-layers are:
— $6 + 4 + 3$, which is not possible because of the pair $(3, 4)$.
Assuming a cuspid of $b_{14}$; the possible middle-layers are:
— $7 + 4 + 3$, which is not possible because of the pair $(3, 4)$.
— $6 + 5 + 3$, which is not possible because of the pair $(3, 5)$.
Assuming a cuspid of $b_{15}$; the possible middle-layers are:
— $8 + 4 + 3$, which is not possible because of the pair $(3, 4)$.
— $7 + 5 + 3$, which is not possible because of the pair $(3, 5)$.
— $6 + 5 + 4$, which is not possible because of the pair $(4, 6)$.
Assuming a cuspid of $b_{16}$; the possible middle-layers are:
— $9 + 4 + 3$, which is not possible because of the pair $(3, 4)$.
— $8 + 5 + 3$, which is not possible because of the pair $(3, 5)$.
— $7 + 6 + 3$, which is not possible because of the pair $(3, 6)$.
— $7 + 5 + 4$, which is not possible because of the pair $(4, 5)$.
Assuming a cuspid of $b_{17}$; the possible middle-layers are:
— $10 + 4 + 3$, which is not possible because of the pair $(3, 4)$.
— $9 + 5 + 3$, which is not possible because of the pair $(3, 5)$.
— $8 + 6 + 3$, which is not possible because of the pair $(3, 6)$.
— $8 + 5 + 4$, which is not possible because of the pair $(4, 5)$.
— $7 + 6 + 4$, which is not possible because of the pair $(4, 6)$.
Assuming a cuspid of $b_{18}$; the possible middle-layers are:
— $11 + 4 + 3$, which is not possible because of the pair $(3, 4)$.
— $10 + 5 + 3$, which is not possible because of the pair $(3, 5)$.
— $9 + 6 + 3$, which is not possible because of the pair $(3, 6)$.
— $9 + 5 + 4$, which is not possible because of the pair $(4, 5)$.
— $8 + 7 + 3$, which is not possible because of the pair $(3, 7)$.
— $8 + 6 + 4$, which is not possible because of the pair $(4, 6)$.
— $7 + 6 + 5$, which is not possible because of the combination of pairs $(5, 6) $ and $(6, 7)$.
It is not hard to continue this logic to find:
$b_{20}$ = $b_{9}$ + $b_{7}$ + $b_{4}$
$b_{9}$ = $b_{6}$ + $b_{2}$ + $b_{1}$
$b_{7}$ = $b_{0}$ + $b_{2}$ + $b_{5}$
$b_{4}$ = $b_{0}$ + $b_{1}$ + $b_{3}$