Deleted $1,2,3,...,9$ from the set, the sum of nine pairwise distinct integers from the remaining is at least $10+11+12+...+18>100$.

Suppose the contradiction. Let $i_1<i_2<...<i_7<i_8$ are eight smallest positive integer that wasn't deleted. Note that this means we've deleted $i_j-j$ numbers, for every $j\in \{ 1,2,...,8\}$, hence $$i_7-7\leq 8\implies i_7\leq 15.$$So, $100-i_1-i_2-...-i_7\geq 100-(15+14+13+...+9)=16$, which is greater than $i_7$. This means we must also delete $100-i_1-i_2-...-i_7$. So, $100-i_1-i_2-...-i_7\neq i_8$. Combine this with the fact that $100-i_1-i_2-...-i_7\geq 16$ and $i_8\leq 16$, we get that either $i_8<16$, or $100-i_1-i_2-...-i_7>16\geq i_8$. The latter case gives us $i_j-j\leq 7$ for all $j\in \{ 1,2,...,8\}$. Hence, both cases lead to $i_8\leq 15$. We, then, apply the same argument to get that we must delete $100-\sum_{j\in J}{i_j}$ where $J$ is a seven-element subset of $\{ 1,2,...,8\}$. Note that all of those eight numbers are distinct and all not smaller than $16$. This means none of $1,2,...,15$ were deleted. So, $i_j=j$ for every $j\in \{ 1,2,...,8\}$. So, we must deleted $100-\sum_{j\in J}{j}$ where $J$ is a seven-element subset of $\{ 1,2,...,8\}$, which consist of $65,66,67,...,72$. Finally, we get that $9+10+11+...+16=100$, contradiction.