Is it possible for long-run variance to be negative or 0? I read that "strictly stationary implies that long-run variance is finite and positive", which means that long-run variance can be 0 or negative. However, according to this, it seems that long-run variance can not be negative. 
So is it possible for long-run variance to be negative or 0?
 A: 
strictly stationary implies that long-run variance is finite and positive

No, that isn't true. Let $X_1, \ldots, X_t$ be iid Cauchy random variables. None of these even have a variance (or mean). Yet the process is stationary. 

which means that long-run variance can be 0 or negative

No variance can ever be negative, because for any random variable $Y$, $(Y-E[Y])^2 \ge 0$ with probability $1$, then its expectation must be non-negative.
Under typical assumptions, it's strictly positive, too. If you look at the formula from the link you posted, 
$$
\lim_{T\to \infty}\text{Var}[\sqrt{T}(\bar{X}_T - \mu)] = \sum_{i=-\infty}^{\infty} \gamma(i)
$$
you might not be able to tell right away because a lot of those terms in the sum might be negative. However, recall that autocovariance functions $\gamma(\cdot)$ are positive definite. So that term has to be positive. The only thing you have to worry about is whether it's $\infty$ or not. But that is usually taken care of by the assumption of absolute summability, or that $\sum_i |\gamma(i)| < \infty$. Finiteness of this implies finiteness of the other.

So is it possible for long-run variance to be negative or 0?

A stationary process can have a long-run variance of $0$, sure. Take a random sample from a distribution that only has probability on $\mu$. Then $\text{Var}(\sqrt{n}(\bar{X}_n - \mu) = 0$ which means its limit is $0$ as well.
A: If you use the popular equation Var(X)=E[X²]-E[X]² then the well-known numerical instabilities can cause the result to be negative if E[x] is larger than the standard deviation. But then the result is clearly bad, because variance is never negative, and negative values cause the standard deviation to be undefined.
While this equation is found in many many textbooks and papers (and is undoubtedly mathematically correct), it is all but reliable with finite floating point numbers as used in computers. So it is okay to use this in proofs, but not in code.
A: Remember that long-run variance is a limit, so it is a little bit different than what one regularly thinks of as variance. For simplicity lets assume $\mu=0$
$$\lim_{T\to\infty}\{\text{Var}[\sqrt{T}\bar{Y}_T]\}=\lim_{T\to\infty}\left\{\text{Var}\left[\frac{1}{\sqrt{T}}\left(\sum_i^T Y_i\right)\right]\right\}.$$
Let $Y$ follow a MA(1) process such that
\begin{align}Y_1&=Z_1-Z_0 \;\;\;\;\; Z_i\sim \mathcal{N}(0,1)\\ Y_t&=Z_t-Z_{t-1}\end{align}
this leads to
$$\text{Var}\left[\frac{1}{\sqrt{T}}\sum_{i}^TY_i\right]=\text{Var}\left[\frac{1}{\sqrt{T}}(Z_0+Z_T)\right]=\frac{2}{T} $$
which in the limit approaches 0.
