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Answer 1

Visual display of the fractions of interest:\[\begin{array}{c|cccccc}
&1&&2&&3&&\cdots&&15&&16&&17\\
\hline
1&\bullet&&\dfrac{1}{2}&&\dfrac{1}{3}&&\cdots&&\dfrac{1}{15}&&\dfrac{1}{16}&&\dfrac{1}{17}\\[1ex]
2&\bullet&&\bullet&&\dfrac{2}{3}&&\cdots&&\dfrac{2}{15}&&\dfrac{2}{16}&&\dfrac{2}{17}\\[1ex]
3&\bullet&&\bullet&&\bullet&&\cdots&&\dfrac{3}{15}&&\dfrac{3}{16}&&\dfrac{3}{17}\\[1ex]
\vdots&\vdots&&\vdots&&\vdots&&\ddots&&\vdots&&\vdots&&\vdots\\[1ex]
15&\bullet&&\bullet&&\bullet&&\cdots&&\bullet&&\dfrac{15}{16}&&\dfrac{15}{17}\\[1ex]
16&\bullet&&\bullet&&\bullet&&\cdots&&\bullet&&\bullet&&\dfrac{16}{17}
\end{array}\]
Summing across the rows boils down to evaluating
\[\sum_{k=1}^{16}k\sum_{n=k+1}^{17}\frac{1}{n}\]and across columns,
\[\sum_{k=2}^{17}\sum_{n=1}^{k-1}\frac{n}{k}\]The second is far easier to work with, as Faulhaber's formula gives
\[\sum_{n=1}^{k-1}n=\frac{k(k-1)}{2}\]so we have
\[\sum_{k=2}^{17}\sum_{n=1}^{k-1}\frac{n}{k}=\sum_{k=2}^{17}\frac{k(k-1)}{2k}=\frac{1}{2}\sum_{k=2}^{17}(k-1)=\frac{1}{2}\sum_{k=1}^{16}k=\frac{16\times17}{2\times2}=68\]

Answer 2

What's the source of this question?