In which case $\mathbb E[X]=\sum _ix_i P[x_i]$ can be $0$ when all $x$'s are not zero ($0$)? Say $X$ is a random variable and $x$'s are realizations of $X$ .
Say , $\mathbb E[X]=\sum _ix_i P[x_i]=0$ . But I do not understand in which case $\mathbb E[X]=\sum _ix_i P[x_i]$  can be $0$ when all $x$'s are not zero ($0$) ?
 A: At least one of the $x_i$ in the support of $X$ must be negative for the mean to be zero, so long as there is non-zero probability of a positive $x_i$ occurring. Otherwise each $x_i P(X=x_i) \geq 0$, and there is at least one $i$ for which $x_i P(X=x_i) > 0$, so $\mathbb{E}(X) = \sum_{i=1}^{n} x_i P(X=x_i) > 0$.
By analogy to mechanics, the mean is the "first moment" of the distribution. Imagine a light see-saw with weights corresponding to the probability masses, $P(X=x_i)$, each placed at a position $x_i$ from the zero point on the line. A negative $x_i$ would be on the left of the zero point, a positive $x_i$ on the right. If the mean is zero, that is saying the centre of mass of all these probability masses is at the zero point — so the see-saw would be perfectly balanced if we placed the pivot there. 

In the example above, I put:
$\Pr(X=x) =  
   \begin{cases}
    2/10  & \text{if } x = -3 \\
    3/10 &\text{if } x = 0 \\
    4/10  & \text{if } x = 1 \\
    1/10  & \text{if } x = 2 \\
    0 & \text{otherwise}
   \end{cases}$
Can you see, physically, why the see-saw will not turn about the pivot, because its clockwise and anticlockwise moments about zero are balanced? Can you use $\mathbb{E}(X)=\sum_i x_i P(X=x_i)$ to show why $\mathbb{E}(X)=0$? Of course, there are very many different distributions I could have chosen. Any distribution which is symmetric about zero, such as the Rademacher distribution in Stephan Kolassa's answer, or the famous standard normal distribution, as an example of a continuous variable, would have worked too. But this example shows some asymmetric distributions also work.
If you only allowed a probability mass at zero, and at least one probability mass on the right hand side (corresponding to a positive $x_i$), but disallowed any masses on the left hand side (negative $x_i$) then there would be a moment about zero turning it clockwise. So we could not balance the see-saw with a pivot at zero, hence zero cannot be the mean. Intuitively this is why, for a zero mean, we need to balance out the positive $x_i$ on the right with at least one negative $x_i$ on the left which can give us the counterbalancing anti-clockwise moment.
Consider the random variable with the following probability mass function. It would not balance about zero (it would tip clockwise if you pivoted it there), but would balance about a pivot at $\mathbb{E}(X)=2$. Note that taking moments about zero finds the first (raw) moment, which by definition is equal to the mean. You can verify that, taking clockwise as positive, the total moment about zero here is indeed $2$. On the other hand, taking moments about the mean finds the first central moment. Now as the distribution balanced about the mean, the total moment is zero: this is a general result, and the first central moment, if it exists, is always zero. (The second central moment, where we square the distance from the mean before summing, is rather more interesting: it is the variance of the distribution.)
\begin{array} {|c|c|c|c|}
\hline
x & 0 & 1 & 3 \\
\hline
\Pr(X=x) & \frac{1}{7} & \frac{2}{7} & \frac{4}{7}\\
\hline
\end{array}

We could transform this distribution to have zero mean by moving the probability masses around. One simple way is to follow Glen_b's suggestion and define $Y=X - \mu_X$. The new mean is $\mathbb{E}(Y)=\mathbb{E}(X - \mu_X) = \mathbb{E}(X) - \mathbb{E}(\mu_X) = \mu_X - \mu_X = 0$ (using linearity of expectations, and that the expectation of a constant is the constant itself). In our case $Y=X-2$: we translate the masses on the see-saw left by two units. After the translation, we have probability masses in both positive and negative positions, so the clockwise and anticlockwise moments about zero can cancel each other out. Note how the centre of mass can be at zero, even with no mass placed there!
\begin{array} {|c|c|c|c|}
\hline
y & -2 & -1 & 1 \\
\hline
\Pr(Y=y) & \frac{1}{7} & \frac{2}{7} & \frac{4}{7}\\
\hline
\end{array}

R code for see-saw plots (pivots added in MS Paint)
seesawPlot <- function(values, masses, xlimits=c(-max(abs(values)), max(abs(values)))) {
    x <- rep(values, times=masses)
    y <- unlist(lapply(masses, function(i){1:i}))
    plot(x, y, ylim=c(0,50), xlim=xlimits, pch=19, col="blue",
        yaxt="n", yaxs="i", frame=F, xlab="", ylab="")
}

values <- c(-3, 0, 1, 2); masses <- c(2, 3, 4, 1)
seesawPlot(values, masses)

values <- c(0, 1, 3); masses <- c(1, 2, 4)
seesawPlot(values, masses)

values <- values - 2
seesawPlot(values, masses, xlimits=c(-3,3))

A: 
But I do not understand in which case $E[X]=∑_ix_iP[x_i]$ can be $0$ when all $x$'s are not zero ($0$)

You can construct any number of such distributions.
e.g. let $Y\sim F_Y$ for some distribution function $F_Y$ with mean $\mu$, in which not all values lie at the mean -- then some will lie above it and some will lie below it.
Let $X=Y-\mu$. Then $E(X)=0$. 
($\text{Var}(X)$ and indeed all central moments will be unaffected by the shift.)

The mass that was below the mean will now lie below 0, and the mass above it will lie above zero.

After pondering the question more closely, I wonder if the question here:

$E[X]=∑_ix_iP[x_i]$ can be $0$ when all $x$'s are not zero ($0$)

is actually "how can the expectation be a value that's not observable"? The answer to that follows directly from the definition of expectation. It's analogous to asking "how can the center of mass of the Pluto-Charon system lie outside either body?" (The same is true of the earth-moon system.)
The center of mass can be in a place where there's no mass at all. It is no more astonishing for probability mass (or probability density) than it is for physical mass.
Handwaving aside, the answer eventually falls back to "because it's defined that way".
If I take a fair six-sided die, the average outcome is 3.5 even though you can never observe it in a single outcome (and most unfair (loaded) six-sided dice also have unobservable means). There's nothing magical about it - the long term average from many tosses can be a value you can't see in a single toss. Indeed, it's acceptable for it to be a value you can't observe in any finite number of tosses (the expectation can be irrational).
A: How about
$$ x_1 = 1, P(x_1)=\frac{1}{2}, \qquad x_2 = -1, P(x_2)=\frac{1}{2} $$
