16. Let $f(x) = \sum_{k=1}^{70} \frac{k}{x-k}$. For all integers $i = 1, \ldots, 70$ we have that $f(x)$ tends to plus infinity as $x$ tends downward to $i$, and $f(x)$ tends to minus infinity as $x$ tends upward to $i$. As $x$ tends to infinity, $f(x)$ tends to 0. Hence it follows that there exist $x_{1}, x_{2}, \ldots, x_{70}$ such that $1 < x_{1} < 2 < x_{2} < 3 < \cdots < x_{69} < 70 < x_{70}$ and $f(x_{i}) = \frac{5}{4}$ for all $i = 1, \ldots, 70$. Then the solution to the inequality is given by $S = \bigcup_{i=1}^{70} (i, x_{i}]$. For numbers $x$ for which $f(x)$ is well-defined, the equality $f(x) = \frac{5}{4}$ is equivalent to
$$p(x) = \prod_{j=1}^{70} (x-j) - \frac{4}{5} \sum_{k=1}^{70} k \prod_{\substack{j=1 \\ j \neq k}}^{70} (x-j) = 0$$
The numbers $x_{1}, x_{2}, \ldots, x_{70}$ are then the zeros of this polynomial. The sum $\sum_{i=1}^{70} x_{i}$ is then equal to minus the coefficient of $x^{69}$ in $p$, which equals $\sum_{i=1}^{70} \left( i + \frac{4}{5} i \right)$. Finally,
$$|S| = \sum_{i=1}^{70} (x_{i} - i) = \frac{4}{5} \cdot \sum_{i=1}^{70} i = \frac{4}{5} \cdot \frac{70 \cdot 71}{2} = 1988$$