What are the minimum and maximum values of variance? I am new to statistics. I am getting my hands dirty on VarianceThreshold. I am having a single dimensional array, containing N values. What's the maximum and minimum values of a variance for any values present in array?
I guess the minimum value will always be non-negative but I don't know about the maximum value. I have googled it but couldn't find a good answer.
 A: It depends... But on what exactly? 
The answer depends on the assumptions that you make. 
1. If I read your question most literally: you know all data values.
In that case, there is no need for bounds (minimum or maximum), as you can simply calculate the variance of the data values in the array with:
$$ \text{var}(\mathbf{x}) = \frac{1}{N} \sum_{i=1}^N (x_i - \bar{x})^2 \, . $$
2. Now, say, you do not know any of the values; only that there are $N$. In other words: you have not seen the sample, but only know the sample size.
Then the answer depends on what you assume about where the sample came from, i.e. the population.
2.1 If you make no assumptions about the population (equivalently: the underlying distribution), then you cannot say anything about an upperbound for the sample variance. 
Take the example of a $t$-distribution with $\nu \le 2$ degrees of freedom. The population variance is infinite, and the sample variance cannot be bounded from above (unless $N = 1$). Why? Because for every sample you provide, I can increase its variance by pushing the minimum and maximum values farther from the mean.
Please note, this same argument holds for a standard normal distribution! Even-though it has population variance equal to $1$, one can create samples with arbitrarily large sample variance.
2.2 If you assume that the population "lives on" bounded support, then Dilip Sarwate's answer will suffice: on support $[0, \, c]$ the sample variance is maximally $c^2 / 4$ (multiplied by $(N-1)/N$ for odd $N$).
P.S. Since the variance is essentially a weighted sum (integral) of non-negative terms (integrand), it is non-negative itself and bounded from below by $0$. I therefore concentrated on the upperbound in my answer.
A: I interpret this question as asking

Given a set of $N$ numbers $x_1, x_2, \ldots,x_N$, what is the minimum and maximum values that the variance $V$, defined as
$$V = \frac 1N \sum_{k=1}^N (x_k-\bar{x})^2 ~~ \text{where}~\bar{x}=\frac 1N \sum_{k=1}^N  x_k$$ can take on?

Well, the minimum value of $V$ is $0$ as Daniel Lopez's comment points out, and it occurs if and only if all the $N$ numbers have the same value.  At the other end, every finite set of real numbers has a (finite) upper bound (call it $b$) and a (finite) lower bound (call it $a$), and
$$V \leq \frac{(b-a)^2}{4} = \left(\frac{b-a}{2}\right)^2 = \left(\frac{\mathcal R}{2}\right)^2\tag{1}$$ where $\mathcal R$ is the range of the set of $N$ numbers. Note that it is not necessary to know the values of $b$ and $a$ separately; we only need the range $\mathcal R = (b-a)$  to calculate the upper bound $(1)$ on $V$.
If $N$ is an even number, there exist sets for which the bound $(1)$ holds with equality: these are sets for which $\frac N2$ of the $x_k$ have value $b$ and the other $\frac N2$ have value $a$. For odd $N$, the bound $\frac{(b-a)^2}{4}$ still applies but cannot be attained with equality if $N>1$. For odd $N>1$, the maximum value is $\frac{(b-a)^2}{4}\frac{N^2-1}{N^2}$ and is attained by a set in which $\frac{N-1}{2}$ of the $x_k$ have value $b$, $\frac{N+1}{2}$ have value $a$, or vice versa. For details, see this answer of mine on math.SE.
In another answer (and the comments on it), @Jim has argued that "A set of $N$ real numbers" tells the listener nothing whatsoever about the set if you don't know what any of the values are, even the minimum or maximum, and so the only completely correct answer is that maximum possible $V$ is unbounded: any other answer (such as mine above or a couple of possible answers suggested by Jim) must be be festooned with caveats that the answer is based on assumptions might be unwarranted. I disagree. Even if a secretive questioner is unwilling to share any details about the set he/she is concerned about, my answer gives the questioner enough information to find the maximum possible value of $V$ for him/herself from very minimal information abut the set: just the range suffices, no need to know even what $N$ is!
EDIT: (by AHK) Corrected the maximum variance for odd $N>1$ and the corresponding choice of $x_i$'s
