$\because$ $\frac{1}{{{x_1}}},\frac{1}{{{x_2}}},\frac{1}{{{x_3}}},….,\frac{1}{{{x_n}}}$ are in $A.P.$

${x_1} = 4\,\,\,\,\,{x_{21}} = 20$

Let $'d'$ be the common difference of this $A.P.$

$\therefore $ its ${21^{st}}$ term $ = \frac{1}{{{x_{21}}}} = \frac{1}{{{x_1}}} + \left[ {\left( {21 – 1} \right) \times d} \right]$

$ \Rightarrow d = \frac{1}{{20}} \times \left( {\frac{1}{{20}} – \frac{1}{4}} \right) \Rightarrow d =  – \frac{1}{{100}}$

Alson  ${x_n} > 50$  (given).

$\therefore \frac{1}{{{x_n}}} = \frac{1}{{{x_1}}} + \left[ {\left( {n – 1} \right) \times d} \right]$

$ \Rightarrow {x_n} = \frac{{{x_1}}}{{1 + \left( {n – 1} \right) \times d \times {x_1}}}$

$\therefore {x_n} = \frac{{{x_1}}}{{1 + \left( {n – 1} \right) \times d \times {x_1}}} > 50$

$ \Rightarrow \frac{4}{{1 + \left( {n – 1} \right) \times \left( { – \frac{1}{{100}}} \right) \times 4}} > 50$

$ \Rightarrow 1 + \left( {n – 1} \right) \times \left( { – \frac{1}{{100}}} \right) \times 4 < \frac{4}{{50}}$

$ \Rightarrow \left( { – \frac{1}{{100}}} \right)\left( {n – 1} \right) <  – \frac{{23}}{{100}}$

$ \Rightarrow n – 1 > 23\,\,\,\,\,\, \Rightarrow n > 24$

Therefore,$n=25$.

$ \Rightarrow \sum\limits_{i = 1}^{25} {\frac{1}{x}}  = \frac{{25}}{2}\left[ {\left( {2 \times \frac{1}{4}} \right) + \left( {25 – 1} \right) \times \left( { – \frac{1}{{100}}} \right)} \right] = \frac{{13}}{4}$