The answer is $2^{2025}-1$; in general, for an upper bound of $2^k$, the maximum $L = 2^{k+1}-1$.

Construction: Let $v_2(x)$ denote the power of $2$ in $x$, and take $a_i = 2^{k-v_2(i)}$. Clearly $a_i \le 2^k$. Moreover, we have:

Lemma: For any $1 \le i \le j \le 2^{k+1}-1$, there exists a unique $i \le x \le j$ such that $v_2(x) = \max_{i \le y \le j} v_2(y)$.

Proof: If both $x$ and $y$ attain the maximum value $v$, this means $x = p \times 2^v$ and $y = q \times 2^v$, where $p$ and $q$ are odd, and without loss of generality assume $p < q$. But then, we have $p < p+1 < q$, so $z = (p+1) \times 2^v$ lies between $x$ and $y$, but $v > 0$, contradicting the maximality of $v$.

Now, for any $1 \le i \le j \le 2^{k+1}-1$, there exists a unique $i \le x \le j$ such that $v_2(x)$ is maximal. This means $v_2(a_x) = k-v$ but $v_2(a_y) > k-v$ for all $y \ne x$, and hence $v_2(\sum s_\ell a_\ell) = k-v$, so it cannot be $0$.

Estimate: Suppose $L \ge 2^{k+1}$. Suppose $a_1, \dots, a_L$ satisfy $a_i \le 2^k$. Let $b_0 = 0$, and recursively define
$$s_i = \begin{cases} +1 & \text{if } b_{i-1} \le 0, \\ -1 & \text{if } b_{i-1} \ge 1. \end{cases}$$
$$b_i = b_{i-1} + s_i a_i.$$

Now, consider the sequence $b_0, b_1, \dots, b_L$. Note that, since $a_i \le 2^k$, if $b_{i-1} \in [-2^k+1, 0]$, then $b_i = b_{i-1}+a_i \in [-2^k+1, 2^k]$; conversely, if $b_{i-1} \in [1, 2^k]$, then $b_i = b_{i-1}-a_i \in [-2^k+1, 2^k]$. Therefore, $b_0$ through $b_L$ consist of $L+1 \ge 2^{k+1}+1$ terms in total, yet there are only $2^{k+1}$ possible values, so there exist $1 \le i \le j \le L$ such that $b_{i-1} = b_j$, that is, $b_j - b_{i-1} = \sum_{\ell \le j} s_\ell a_\ell = 0$. Hence there does not exist a sequence of length greater than $2^{k+1}-1$ satisfying the condition.