Answer: $2^{2024}-1$ .

Construction:

Let's define sequence $A_0$ as just number $1$ . For convenience we will mean sequence $kB$ as sequence $B$ , every term of which is increased by $k$ times. Now, define other sequences recurrently:

$A_{n+1}=2A_n, 1, 2A_n$

For example, $A_1=212$ and $A_2=4241424$ .

Now, this construction works because there is only one $1$ meaning that by parity if it is included in the $s_i, ..., s_j$ the sum can't be zero. On the other hand, the parts on the $RHS$ and $LHS$ of $1$ are working by induction.

Bound:

We will prove by induction a stronger statement, that if we have a sequence bigger than that, we can find a sequence of '+' and '-' such that the amount of '+' equals to the amount of '-'. Suppose that we have a sequence of length greater or equal to $2^{n+1}-1$ if $a_i \le 2^n$ . Then, we may assume that the sequence has exactly $2^{n+1}$ terms, say $a_1, a_2, ..., a_{2^{n+1}}$ . We then define $b_i=\mid a_{2i} - a_{2i-1} \mid$ . This gives us $2^n$ numbers that are between $0$ and $2^n$ . Let $c_i=b_i-2^{n-1}$ . By induction, we can assume that this sequence $c_i$ has a sequence of '+' and '-' that we searched for. Then the sequence $b_i$ can be put into $0$ yet we discard our statement about the amount of '+' and '-'. However, because of $b_i$ 's definition, when we put signs next to $a_{2i}$ and $a_{2i-1}$ these signs will be different (that form $b_i$ ). Therefore the amount of '+' and '-' is again equal so induction is complete.