smc

Suppose that there are $n$ positive integers in the set initially, so their sum is $\frac{n(n+1)}{2}$ by arithmetic series. The average of the remaining numbers is minimized when $n$ is erased, and is maximized when $1$ is erased. It is clear that $n>1.$ We write and solve a compound inequality for $n:$ \begin{alignat}{8} \frac{\frac{n(n+1)}{2}-n}{n-1} &\leq 35\frac{7}{17} &&\leq \frac{\frac{n(n+1)}{2}-1}{n-1} \\ \frac{n(n+1)-2n}{2(n-1)} &\leq 35\frac{7}{17} &&\leq \frac{n(n+1)-2}{2(n-1)} \\ \frac{n^2-n}{2(n-1)} &\leq 35\frac{7}{17} &&\leq \frac{n^2+n-2}{2(n-1)} \\ \frac{n(n-1)}{2(n-1)} &\leq 35\frac{7}{17} &&\leq \frac{(n+2)(n-1)}{2(n-1)} \\ \frac{n}{2} &\leq 35\frac{7}{17} &&\leq \frac{n+2}{2} \\ n &\leq 70\frac{14}{17} &&\leq n+2 \\ 68\frac{14}{17} &\leq \hspace{3mm} n &&\leq 70\frac{14}{17}, \end{alignat} from which $n$ is either $69$ or $70.$ Let $x$ be the number that is erased. We are given that $\frac{\frac{n(n+1)}{2}-x}{n-1}=35\frac{7}{17},$ or $$\frac{n(n+1)}{2}-x=35\frac{7}{17}\cdot (n-1).(★)$$ If $n=70,$ then $(★)$ becomes $2485-x=\frac{41538}{17},$ from which $x=\frac{707}{17},$ contradicting the precondition that $x$ is a positive integer. If $n=69,$ then $(★)$ becomes $2415-x=2408,$ from which $x=\boxed{7}.$ Remark From $(★),$ note that the left side must be an integer, so must be the right side. It follows that $n-1$ is divisible by $17,$ so $n=69.$