Combination Summation problem 
I obtained $np/2$ which has a similar form as q but I'm unable to prove that it is equal to $q$.
 A: As a first step, we have better to simplify and pass from the double sums to simple ones.
Let's put
$$
s(n) = \sum\limits_{1\, \le \,k\, \le \,n} {\frac{1}{{\left( \begin{array}{c}  n \\  k\\
 \end{array} \right)}}}  = \frac{1}{{n!}}\sum\limits_{1\, \le \,k\, \le \,n} {k!\left( {n - k} \right)!} 
$$
Then
$$
\begin{array}{l}
 \sum\limits_{1\, \le \,k\, < \,j\, \le \,n} {\frac{1}{{\left( \begin{array}{c}  n \\   k \\ 
 \end{array} \right)}} + \frac{1}{{\left( \begin{array}{c}  n \\   j \\ 
 \end{array} \right)}}}  + \sum\limits_{1\, \le \,j\, < \,k\, \le \,n} {\frac{1}{{\left( \begin{array}{c}  n \\   k \\ 
 \end{array} \right)}} + \frac{1}{{\left( \begin{array}{c}  n \\   j \\ 
 \end{array} \right)}}}  + 2\sum\limits_{1\, \le \,k\, \le \,n} {\frac{1}{{\left( \begin{array}{c}  n \\   k \\ 
\end{array} \right)}}}  =  \\ 
  = 2p(n) + 2s(n) =  \\ 
  = \sum\limits_{\begin{array}{*{20}c}    {1\, \le \,k\, \le \,n}  \\    {1\, \le \,j\, \le \,n}  \\
\end{array}} {\left( {\frac{1}{{\left( \begin{array}{c}  n \\   k \\ 
 \end{array} \right)}} + \frac{1}{{\left( \begin{array}{c}  n \\   j \\ 
 \end{array} \right)}}} \right)}
  = \sum\limits_{1\, \le \,k\, \le \,n} {\sum\limits_{1\, \le \,j\, \le \,n} {\left( {\frac{1}{{\left( \begin{array}{c}  n \\   k \\ 
 \end{array} \right)}} + \frac{1}{{\left( \begin{array}{c}  n \\   j \\ 
 \end{array} \right)}}} \right)} }  =  \\ 
  = \sum\limits_{1\, \le \,k\, \le \,n} {\left( {\frac{n}{{\left( \begin{array}{c}  n \\   k \\ 
 \end{array} \right)}} + s(n)} \right)}  = 2n\,s(n)\quad  \Rightarrow  \\ 
  \Rightarrow \quad p(n) = \left( {n - 1} \right)\,s(n) \\   \end{array}
$$
Similarly, putting
$$
r(n) = \sum\limits_{1\, \le \,k\, \le \,n} {{k \over {\binom{n}{k}}}} 
$$
we get
$$
\eqalign{
 & \sum\limits_{1\, \le \,k\, < \,j\, \le \,n} {
{k \over {\binom {n}{k}}} + {j \over {\binom {n}{j}}}} 
 + \sum\limits_{1\, \le \,j\, < \,k\, \le \,n} {{k \over {\binom {n}{k}}}
 + {j \over {\binom {n}{j}}}} 
 + 2\sum\limits_{1\, \le \,k\, \le \,n} {{k \over {\binom {n}{k}}}}  =   \cr  
 &  = 2q(n) + 2r(n) =   \cr 
 &  = \sum\limits_{\matrix{   {1\, \le \,k\, \le \,n}  \cr  
  {1\, \le \,j\, \le \,n}  \cr  } } {\left( {{k \over {\binom {n}{k}}}
 + {j \over {\binom {n}{j}}}} \right)} 
 = \sum\limits_{1\, \le \,k\, \le \,n} {\sum\limits_{1\, \le \,j\, \le \,n} {
\left( {{k \over {\binom {n}{k}}}
 + {j \over {\binom {n}{j}}}} \right)} }  =   \cr  
 &  = \sum\limits_{1\, \le \,k\, \le \,n} {\left( {n{k \over {\binom {n}{k}}} + q(n)} \right)}  = 2n\,r(n)
 \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad q(n) = \left( {n - 1} \right)\,r(n) \cr} 
$$
We can then work on $s(n),\, r(n)$ to obtain
$$
\eqalign{
  & r(n) + s(n) = \sum\limits_{1\, \le \,k\, \le \,n} {{{k + 1} \over {\binom{n}{k}}}}  =   \cr 
  &  = {1 \over {n!}}\sum\limits_{1\, \le \,k\, \le \,n} {\left( {k + 1} \right)!\left( {n - k} \right)!}  =   \cr 
  &  = \left( {n + 1} \right)\left( {{1 \over {\left( {n + 1} \right)!}}\sum\limits_{1\, \le \,k\, \le \,n} {\left( {k + 1} \right)!\left( {n + 1 - \left( {k + 1} \right)} \right)!} } \right) =   \cr 
  &  = \left( {n + 1} \right)\left( {{1 \over {\left( {n + 1} \right)!}}\sum\limits_{2\, \le \,k\, \le \,n + 1} {k!\left( {n + 1 - k} \right)!} } \right) =   \cr 
  &  = \left( {n + 1} \right)\left( {s(n + 1) - {{n!} \over {\left( {n + 1} \right)!}}} \right) = \left( {n + 1} \right)s(n + 1) - 1 \cr} 
$$
and
$$
\eqalign{
  & s(n + 1) = \sum\limits_{1\, \le \,k\, \le \,n + 1} {{1 \over {\left( \matrix{
  n + 1 \cr 
  k \cr}  \right)}}}  =   \cr 
  &  = {1 \over {\left( {n + 1} \right)!}}\sum\limits_{1\, \le \,k\, \le \,n + 1} {k!\left( {n + 1 - k} \right)!}  =   \cr 
  &  = {1 \over {\left( {n + 1} \right)!}}\left( {\sum\limits_{1\, \le \,k\, \le \,n} {k!\left( {n + 1 - k} \right)!}  + \left( {n + 1} \right)!} \right) =   \cr 
  &  = 1 + {1 \over {\left( {n + 1} \right)!}}\sum\limits_{1\, \le \,k\, \le \,n} {\left( {n + 1 - k} \right)k!\left( {n - k} \right)!}  =   \cr 
  &  = 1 + {1 \over {\left( {n + 1} \right)!}}\left( {\left( {n + 2} \right)\sum\limits_{1\, \le \,k\, \le \,n} {k!\left( {n - k} \right)!}
  - \sum\limits_{1\, \le \,k\, \le \,n} {\left( {k + 1} \right)!\left( {n + 1 - \left( {k + 1} \right)} \right)!} } \right) =   \cr 
  &  = 1 + {1 \over {\left( {n + 1} \right)!}}\left( {\left( {n + 2} \right)n!s(n)
 - \sum\limits_{1\, \le \,k\, \le \,n} {\left( {k + 1} \right)!\left( {n + 1 - \left( {k + 1} \right)} \right)!} } \right) =   \cr 
  &  = 1 + {1 \over {\left( {n + 1} \right)!}}\left( {\left( {n + 2} \right)n!s(n) - \sum\limits_{2\, \le \,k\, \le \,n + 1} {k!\left( {n + 1 - k} \right)!} } \right) =   \cr 
  &  = 1 + {1 \over {\left( {n + 1} \right)!}}\left( {\left( {n + 2} \right)n!s(n) - \left( {n + 1} \right)!s(n + 1) + n!} \right) =   \cr 
  &  = 1 + {{n + 2} \over {n + 1}}s(n) - s(n + 1) + {1 \over {n + 1}}\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad s(n + 1) = {{n + 2} \over {2\left( {n + 1} \right)}}\left( {1 + s(n)} \right) \cr} 
$$
So, the final result is
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  p(n) = \left( {n - 1} \right)\,s(n) \hfill \cr 
  q(n) = \left( {n - 1} \right)\,r(n) \hfill \cr 
  r(n) + s(n) = \left( {n + 1} \right)s(n + 1) - 1 \hfill \cr 
  s(n + 1) = {{n + 2} \over {2\left( {n + 1} \right)}}\left( {1 + s(n)} \right) \hfill \cr} 
 \right.\quad  \Rightarrow \quad \left\{ \matrix{
  r(n) = {n \over 2}\left( {s(n) + 1} \right) \hfill \cr 
  p(n) = \left( {n - 1} \right)\,s(n) \hfill \cr 
  q(n) = {n \over 2}\left( {p(n) + \left( {n - 1} \right)} \right) \hfill \cr}  \right.
 }$$
which is not as expected.
In fact, you should define the sums to start from zero
$$ \bbox[lightyellow] {  
\eqalign{
  & \matrix{
   \matrix{
  p^ *  (n) = \sum\limits_{0\, \le \,k\, < \,j\, \le \,n} {{1 \over {  \binom{n}{k} }} 
 + {1 \over { \binom{n}{j}  }}}  =  \hfill \cr 
    = \sum\limits_{0\, < \,j\, \le \,n} {\left( {1 + {1 \over {\binom{n}{j}}}} \right)}
   + \sum\limits_{1\, \le \,k\, < \,j\, \le \,n} {{k \over {\binom{n}{k} }}
 + {j \over {\binom{n}{j}}}}  =  \hfill \cr 
   = n + s(n) + p(n) \hfill \cr}  & {\;\;\;} & \matrix{
  q^ *  (n) = \sum\limits_{0\, \le \,k\, < \,j\, \le \,n} {{k \over {\binom{n}{k} }} 
 + {j \over {\binom{n}{j}}}}  =  \hfill \cr 
   = \sum\limits_{0\, < \,j\, \le \,n} {\left( {{j \over {\binom{n}{j}}}} \right)}
  + \sum\limits_{1\, \le \,k\, < \,j\, \le \,n} {{k \over {\binom{n}{k}}} 
  + {j \over {\binom{n}{j}}}}  =  \hfill \cr 
   = r(n) + q(n) \hfill \cr}   \cr 
 }   \cr 
  &  \cr} 
 }$$
to obtain
$$ \bbox[lightyellow] {  
q^ *  (n) = {n \over 2}p^ *  (n)
 }$$
