Adapting the analysis from this answer
If two people are the same age we are done, so assume everyone is a different age.
The number of non-empty subsets will be $2^{10}-1=1023$. We can immediately see this is greater than the maximum feasible age sum $\sum_{91}^{100}i = 955$ so there must be some duplicate sums.
Clearly there is room to raise the age limit. This analysis allows us to get to $\color{red}{106}$ since $\sum_{97}^{106}i = 1015<1023$ but we can do better.
Consider the youngest person, with an age of $y$ - this is the lower limit of possible age sums. The upper limit with the orginal age limit is then at most $y+\sum_{92}^{100}i = y+864$. There are thus $864$ or fewer different age sums available for different groups and we can extend the limit by $\lfloor({1023-864})/({9})\rfloor = 17$ years to $\color{red}{117}$.
We're not done...
With distinct ages, the full set sum is clearly not going to match any other subset, and neither is any subset with only one missing person unless that person is the oldest. So in a group of $n$ people we can discard $n$ subsets and reduce the highest possible total by the age limit. So for $10$ people, considering the $1013$ groups defined under this process we can impose an age limit of $\color{red}{130}$ since the feasible totals range of our chosen subsets is $y$ to $y+\sum_{122}^{129}i = y+251\cdot 4 = y+1004$ for $1005$ possible totals, allowing the pigeonhole argument again.
Finally note that this age limit is likely an underestimate of the maximum possible. Leaving only one "young end" age can produce huge gaps in the possible age sums (fewer holes for our pigeons). So it is almost certain that the highest possible age limit is still higher.
