Showing that $S=\frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{1000} \gt 1$ 
If $$S=\frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{1000}$$ then
  $$S\gt 1,$$

but how?
I understood that there are $451$ pair of terms. So clubbed two terms together.
$\frac{1}{100}+\frac{1}{1000}+\frac{1}{101}+\frac{1}{999}.....$
But I am not able to solve it further. Such a tricky question for me.

There are $901$ terms in $S$
Consider the first $100.$  The smallest term of this set is $\frac 1{199}$
The sum of the first $100$ terms is greater than $\frac 12$
Now consider the next $200$ terms.
the smallest term in this set is $\frac {1}{399}$
and the sum of that subset is greater that $\frac 12$
and you are done.  And there are still $601$ you haven't even considered yet.
The harmonic series:
$S = \sum_\limits{n=1}^\infty \frac 1n = 1 + \frac 12 + \frac 13 + \frac 14 \cdots\\
\frac 13 + \frac 14 > \frac 1{4} + \frac 1{4} = \frac 12\\
\frac 15 + \frac 16 + \frac 17 + \frac 18 > \frac 18 + \frac 18 + \frac 18 + \frac 18 = \frac 12\\
$
And we can break infinite series into (infintely many) finite sub-series, and each sub-series sums to something that is greater than $\frac 12$
$S > 1+\frac 12 + \frac 12 + \frac 12 \cdots$ and as we add more terms $S$ marches off to infinity (albeit slowly)
For you problem I have employed a very similar tactic.  I have found subsets of your series, each of which sum to something greater than $\frac 12$
