$ \mathbb{P}( \xi_n > \frac{n}{2}) \geq \delta. $, where $\xi_n$ denotes the number of $X_i$s for which $ \sqrt{n} \leq i \leq n$ and $X_i \geq 1$ Suppose that  $X_1, X_2, \ldots $ are mutually independent and $X_i$ has the following density function:
$$
f_i(x)= 
\begin{cases}
\frac{|x|}{i^2} , \ \  0 \leq |x| \leq  i, \ i=1,2, \ldots \\
0, \ \ \text{sonst}
\end{cases}
$$
Then it is clear, that
$$
\mathbb{P}(X_i \geq 1)= \frac{1}{2} - \frac{1}{(2i^2)},  \ \  i=1,2,\ldots
$$
Denote by $\xi_n$ the number of $X_i$s for which $ \sqrt{n} \leq i \leq n$ and $X_i \geq 1$. Why follows from this, by an application of the central limit theorem, that for some $ \delta \in (0,1/2)$ we have:
$$
\mathbb{P}( \xi_n > \frac{n}{2})  \geq \delta.
$$
for $n$ sufficently large ?

 A: Let (by $\sum_i$ we mean $\sum_{i = [\sqrt{n}]}^n$)$$\xi_n = \sum_{i=[\sqrt{n}]}^n 1_{\{X_i \ge 1\}} =: \sum_{i} Y_i$$
That is $\xi_n$ is the number of random variables $X_i: i \in \{[\sqrt{n}],...,n\}$ such that $X_i \ge 1$ and $Y_i = 1_{\{X_i \ge 1\}}$.
Note that $\mathbb E[Y_i] = \mathbb P(X_i \ge 1) = \frac{1}{2} - \frac{1}{2i^2}$.
So that   $$\mathbb P( \xi_n \ge \frac{n}{2}) = \mathbb P(\sum_{i}Y_i - \sum_i \mathbb E[Y_i] \ge \frac{n}{2} -(\frac{n-[\sqrt{n}]}{2}) + \sum_i \frac{1}{2i^2}) $$
Which is the same as $$\mathbb P( \frac{\sum_i Y_i - \mathbb E[Y_i]}{\sqrt{n}} \ge \frac{1}{2} + \frac{1}{2\sqrt{n}}\sum_i \frac{1}{i^2})$$
Note that $Var(Y_i) = \mathbb P(Y_i \ge 1) - (\mathbb P(Y_i \ge 1))^2 = \frac{1}{4} - \frac{1}{4i^4} $, so $\sum_i Var(Y_i) = (n-[\sqrt{n}])\frac{1}{4} -\sum_i \frac{1}{4i^4} $
The latter goes to $0$ since that series converge.
We can then write:
$$ \mathbb P(2\frac{\sum_i(Y_i - \mathbb E[Y_i])}{\sqrt{n}}  \ge 1 + \frac{1}{\sqrt{n}}\sum_i \frac{1}{i^2}) $$
It isn't hard to check that this sequence on the left satisfies Linderberg CLT condition (note that $|Y_i - \mathbb E[Y_i]| \ge \varepsilon \frac{\sqrt{n}}{2}$ does not hold for any $i \in  \{[\sqrt{n}],...,n\}$ when $n$ is big enough).
Hence on the left we have convergence to $\mathcal N(0,1)$ in distribution. So in the limit our probability is equivalent to: (note that CDF of normal is continuous)
$$\mathbb P(\mathcal N(0,1) \ge 1) $$ which is clearly greater than some $2\delta \in (0,1)$. But since the limit is greater than $2\delta$, starting from some $n$, your sequence must be greater than $\delta \in (0,\frac{1}{2})$
