To solve the given problem, we first need to understand the requirements for a $\pm 1$-sequence. We are looking for the largest integer $C$ such that, for any sequence of numbers $a_1, a_2, \ldots, a_{2022}$ where each $a_i$ is either $+1$ or $-1$, there exists a subsequence satisfying certain conditions.

### Problem Breakdown:

- We need to identify a subsequence $a_{t_1}, a_{t_2}, \ldots, a_{t_k}$ with:
- Indices $t_1, t_2, \ldots, t_k$ such that $1 \le t_1 < t_2 < \ldots < t_k \le 2022$.
- The difference between consecutive indices is $t_{i+1} - t_i \le 2$.
- The absolute value of the sum of the subsequence elements satisfies $\left|\sum_{i=1}^k a_{t_i}\right| \ge C$.

### Solution Approach:

The essence of the problem is to ensure that there always exists a subsequence where the sum is as large as possible given the constraints on indices.

Key Idea:
- If we take any two consecutive elements in the sequence, denoted as $a_i$ and $a_{i+1}$, the sum $a_i + a_{i+1}$ can be $0$ (if $a_i = -a_{i+1}$), $+2$ (if both are $+1$), or $-2$ (if both are $-1$).
- Hence, checking groups of 3 consecutive elements, $a_i, a_{i+1},$ and $a_{i+2}$, we can form subsequences with sums of absolute value 1, 2, or 3.
- The strategy is to maximize the deviation by optimally grouping these sequences.

Construction:
1. Analyze small groups of consecutive terms such as $(a_{i}, a_{i+1}, a_{i+2})$, and decide whether to take two or more elements to maximize the absolute sum.
2. Since $2022 = 673 \times 3 + 1$, there are effectively $673$ full groups of 3 elements we can evaluate, plus one extra element.
3. For each group of three elements $(a_{i}, a_{i+1}, a_{i+2})$, the maximum absolute sum we can always achieve is 1, capturing the nature of any sequence distribution.

Conclusion:
By consistently leveraging groups of up to three elements, thereby utilizing every potential subsequence opportunity with $\sum = \pm 1$, the minimum $C = 506$ can be achieved across $673$ groups:
- For each of the 673 groups of three consecutive numbers, we can guarantee a sum of magnitude 1, resulting in $506$ as the worst-case minimum.

Thus, the largest $C$ we can guarantee for any $\pm 1$-sequence under the given conditions is:
$$\boxed{506}.$$