Solved with L567. $\textbf{Claim:}$ $A < 2$ does not satisfy the condition $\textbf{Proof)}$ Notice that for all sufficiently large $n \in \mathbb{N}$, $$A = x_{n+1} + \frac{1}{x_n} \geq 2 \sqrt{\frac{x_{n+1}}{x_n}}$$by AM-GM and consequently if we let $tA^2 = 4$ for some $t > 1$, then $$x_n > tx_{n+1}$$and therefore we have that $$\frac{x_n}{t^k} > x_{n+k}$$meaning that $x_{n+k}$ will reach arbitrarily small positives, yet, $$x_n > \frac{1}{A}$$for all sufficiently large $n \in \mathbb{N}$ which is a contradiction. $\blacksquare$ $\textbf{Claim:}$ $A \geq 2$ satisfies the condition $\textbf{Proof)}$ Assume that we have at least one negative term in the sequence because otherwise there are clearly only positive terms, then notice that at the next term $x_n > A$. Now, take some $t \in \mathbb{R^{+}}$ such that $A = t + \frac{1}{t}$ which clearly exists as $A \geq 2$. Then notice that if $x_n > A > t$, then $$x_{n+1} = A - \frac{1}{x_n} > t$$and consequently $x_{m+i} > 0$ for all $i \geq 0$, as desired. $\blacksquare$

We will start by observing that $A$ can't be negative,otherwise atleast one of any 2 consecutive term will be negative. Claim: If $A\in(0,2)$ then any sequence satisfying the given recurrence relation has infinitely many negative terms. Proof. FTSOC, assume $\forall n\ge N, x_n>0$ for some $N\in \mathbb N$.Then $\{x_n\}_{n=N}^{\infty}$ must be a decreasing sequence.If not assume for some $n\ge N, x_{n+1}\ge x_n$.Then $$2>A=x_{n+1}+\frac{1}{x_{n}}\ge x_n+\frac{1}{x_n}\ge 2$$Which is certainly impossible. So $\{x_n\}_{n=N}^{\infty}$ is a decreasing sequence bounded below by 0.Hence it has a limit, say $L$.$L$ can't be zero,otherwise for some large $n, \frac{1}{x_n}$ become greater than $A$.So $L>0$ Pick any $\varepsilon>0$.Since $x_n\to L$, there is $K$ such that for all $n\ge K, L\le x_n< \L+\varepsilon$. But then, $$2>A=x_{n+1}+\frac{1}{x_{n}}\ge L+\frac{1}{L+\varepsilon}$$Since we can take $\varepsilon$ arbitrary small, the last inequality can't hold for AM-GM reason.$\blacksquare$ Now we will prove that for any $A\ge 2$, and for any sequence satisfying the recurrence relation, atmost one term can be negative. Indeed,if $x_n<0$ then $x_{n+1}>A\ge 2$.We claim that if any term is greater than equal to 1 then all the next terms are greater than equal to 1. Indeed, if $x_i\ge 1$ then, $$x_{i+1}=A-\frac{1}{x_i}\ge 2-1=1$$Inductively, we are done.$\blacksquare$

Firstly, consider $A<2$. We claim that this does not satisfy the problem. For the sake of contradiction assume we have $N$ such that for all $n\geq N$, $x_n>0$. Thus considering $n\geq N$, $$x_{n+1}=A-\frac{1}{x_n}<\frac{2x_n-1}{x_n}\leq x_n,$$which means that $\{x_n\}_{n=N}^{\infty}$ is decreasing. Note that $\{x_n\}_{n=N}^{\infty}$ is bounded below as otherwise we would have some $n$ with $x_n<0$. Hence, $\{x_n\}_{n=N}^{\infty}$ has a limit, call that $L$. However, note that $$L^2=AL-1<2L-1\implies (L-1)^2<0.$$We obtain a desired contradiction. Now we consider $A\geq 2$. We claim that this does satisfy the problem. Note that if $x_n\geq 1$, then so is $$x_{n+1}=A-\frac{1}{x_n}\geq 2-1=1.$$Now we claim that there is a $n$ such that $x_n\geq 1$, which implies that for all $m\geq n$, we have $x_m\geq 1$. If we have negative $x_n$, then $x_{n+1}>2\geq 1$. Thus, either we have negative, which implies the next $x_n$ positive or we have all $x_n$ positive, we are done.

The sequence $ x_ {n} $ is sometimes just a constant. Very nice problem.Answer:$A\ge 2$

We claim that $\boxed{ A \geq 2}$ works $\textit{case 1:}$ $A\geq 2$ Note the following little obervation: if for any $i$, $x_i$ is negative, it follows that $x_{i+1}$ is bigger than $A$ and therefore necessarely positive. Therefore we may assume that wlog, the first term of the sequence is positive, consider now two subcases: $\textit{case 1.1:}$ $x_1 \geq A$ This assumption yields that $1.5 \leq x_2 \leq 2$, the important part here is that $1<x_2$. Because $1<x_2$, we arrive at the conclusion that $1>\frac{1}{x_2}$. But that argument can easily be extended inductively, therefore no term will ever go below $1$, and consequently ever be negative $\textit{case 1.2:}$ $x_1 < 1$ (note that the case $1 \leq x_1 \leq 2$ trivially follows from above) In this case, the most important note is just that if the sequence ever contains a negative term, the subsequent term is forcefully bigger than $A$, leading us into case $1.1$ $\textit{case 2}$ $0 \leq A < 2$ In this case, choose $x_1 = 1$. By this, $\frac{1}{x_1} < \frac{1}{x_2}$, which means the $x_i$ will grow smaller, and their reciprocals larger, until one of them becomes forcefully negative. Once it has become negative, it will bounce back to positive, but will continue to decrease, until the same procedure will happen again. This is mainly due to the fact that generally the inequality described in case $1.1$ can't be fulfilled since $A-1 < 1$ Now notice that actually, $A$ can be negative. But this is basically just mirroring everything that happens and flipping the sign, so there can't be a solution