, 2022

Solution:
Let the answer be $M$. If $M>1011$, there would exist two consecutive positive terms $x_{k}, x_{k+1}$ which contradicts the assumption that $x_{k}+\frac{1}{x_{k+1}}<0$. Thus, $M \leq 1011$. If $M=1011$, then the $2022 x_{i}$ s must alternate between positive and negative. WLOG, assume $x_{2 k-1}>0$ and $x_{2 k}<0$ for each $k$. Then, we have
$$\begin{aligned} & x_{2 k-1}+\frac{1}{x_{2 k}}<0 \Longrightarrow\left|x_{2 k-1} x_{2 k}\right|<1, \\ & x_{2 k}+\frac{1}{x_{2 k+1}}<0 \Longrightarrow\left|x_{2 k} x_{2 k+1}\right|>1 . \end{aligned}$$
Multiplying the first equation over all $k$ gives us $\prod_{i=1}^{2022}\left|x_{i}\right|<1$, while multiplying the second equation over all $k$ gives us $\prod_{i=1}^{2022}\left|x_{i}\right|>1$. Thus, we must have $M<1011$.
$M=1010$ is possible by the following construction:
$$1,-\frac{1}{2}, 3,-\frac{1}{4}, \ldots, 2019,-\frac{1}{2020},-10000,-10000$$