What is the smallest $2l$ for which the most efficient sequence of numbers is possible? So I'm searching for the a sequence of numbers such that when any $2$ are picked their sum will add up to any even number $2m \geq 2l$ where $2l$ is the smallest even number thier sum is equal to. Then, what is the smallest $2l$ for which the most efficient sequence is possible? To define the most efficient sequence is possible we use the following. Let,
$$S = a_1 + a_2 + a_3+\dots +a_n$$
where $a_1 < a_2 < \dots < a_n$ and are elements of our sequence. Then:
$$ (2 n -1) S = \sum_{i \leq j \leq N} (a_i + a_j)$$
Now we know the sum of these $a_i + a_j$ should yield an even number $ \geq 2 l$. Let us assume that each even number only occurs once. Then:
$$ (2 n -1) S = \sum_{r=l}^{a_n} 2r = a_n(a_n +1) - l(l+1)$$
Diving by $2n-1$:
$$ S  = \frac{a_n(a_n +1)}{2n-1} - \frac{l(l+1)}{2n-1}$$
Hence,
$$ a_n = \frac{a_n(a_n +1)}{2n-1} - \frac{l(l+1)}{2n-1} - \frac{a_{n-1}(a_{n-1} +1)}{2n-3} + \frac{l(l+1)}{2n-3} $$
When $l$ is odd it makes sense to assume $a_1 = 2l$ if $l$ is odd. But for even $l$ $a_1$ can also be $l$. However, all of this assumes the existence of the "the most efficient sequence" and thus we circle back to what is the smallest $2l$ for which the most efficient sequence is possible?
 A: This answer assumes that "the most efficient sequence" $\{a_n\}$ is a sequence satisfying

*

*$a_1\lt a_2\lt \cdots\lt a_n$


*$\{a_i+a_j\mid 1\le i\le j\le n\}=\bigg\{2a_1,2(a_1+1),\cdots, 2\bigg(a_1+\dfrac{n(n+1)}{2}-1\bigg)\bigg\}$

This answer proves that the most efficient sequence exists if and only if $n\le 2$.
Proof :

*

*For $n=1$, the most efficient sequence exists.


*For $n=2$, the most efficient sequence exists since $a_2=a_1+2$ works.


*For $n=3$, suppose that the most efficient sequence exists.Then, since $$a_1+a_1\lt a_1+a_2\lt a_2+a_2,$$
$$a_1+a_2\lt a_1+a_3\lt a_2+a_3\lt a_3+a_3,$$
$$a_3+a_3=2a_1+10\tag1$$we have to have
$$a_1+a_2=2a_1+2\implies a_2=a_1+2$$from which we get $a_2+a_2=2a_1+4$.So, we have to have $$a_1+a_3=2a_1+6\implies a_3=a_1+6$$which contradicts $(1)$.


*For $n\ge 4$, suppose that the most efficient sequence exists.Then, since $$a_1+a_1\lt a_1+a_2\lt a_2+a_2,\qquad a_1+a_2\lt a_1+a_3\lt a_2+a_3\lt a_3+a_3,$$
$$a_1+a_3\lt a_1+a_4$$we have to have
$$a_1+a_2=2a_1+2\implies a_2=a_1+2$$from which we get $a_2+a_2=2a_1+4$.So, we have to have $$a_1+a_3=2a_1+6\implies a_3=a_1+6$$from which we get $a_2+a_3=2a_1+8$ and $a_3+a_3=2a_1+12$.So, we have to have$$a_1+a_4=2a_1+10\implies a_4=a_1+10$$implying $a_2+a_4=a_3+a_3$, which contradicts that $\{a_n\}$ is the most efficient sequence.$\quad\blacksquare$
