A question related to Pigeonhole Principle 
In a room there are $10$ people, none of whom are older than $60$, but each of whom is at least $1$ year old. Prove that one can always find two groups of people (with no common person) the sum of whose ages is the same.

My approach:  There are $2^{10}=1024$ subsets, $1023$ non-empty subsets. Therefore there are $1023$ sums of ages and each sum is between $1$ and $600$. Then there are $600$ possible values, but $1023$ sums. Therefore at least two of them must be equal, i.e. there exist different subsets $\{P_{i1}, \ldots, P_{in}\}$ and $\{P_{j1}, \ldots, P_{jn}\}$ such that the sum of the ages agree. Now take out the people present in both subsets.

Can $10$ people be replaced by a smaller number?

I guess, it cannot. For example if there were to be $9$ people, then I would have $2^9-1 = 511$ proper subsets and since now I have $9\cdot 60=540$ possible totals, it is not guaranteed that there exists two disjoint groups of people such that the sum of whose ages are the same.
Am I right?
 A: Here is a (non-exhaustive) proof that a 9-person set doesn't exist. It looks at a more restrictive set, to greatly reduce the number of pigeonholes.
Let the ages of the people be (WLOG) $1 < a_1 < a_2 < \ldots a_9 \leq 60$. They have to be distinct, otherwise we are done.
Consider all subsets with at least 1 people. There are $2^9 - 1 = 511$ such sets. These are our pigeons, previously 512. The difference between the biggest and the smallest sum of ages is $a_2 + a_3 + \ldots a_9 \leq 452$. As such, the sum of ages can take on at most 453 different values. These are our pigeonholes, previously 540. Hence, by the Pigeonhole principle, there are 2 sets with the same sum. Now take the set difference, to ensure that we get 2 groups with no common people.
This approach does not extend to showing that a 8-person set doesn't exist, as claimed by Ilmari.
A: For the $10$ part you are fine.  For the $9$ part, you haven't proven that it can be done, just that this approach isn't sufficient to rule it out.  One way to finish the $9$ part is to display a set of $9$ numbers that you can't find such a set of subsets.  After a bit of searching I haven't found one.
A: As Ross and others have noted, your argument for 10 people is fine.  To show that it's not possible to find two such groups out of 9 or fewer people, you should either exhibit a 9-person set that does not have two such subsets, or at least somehow prove that such a 9-person set exists.
Unfortunately, according to a brute force computer search I ran, such a counterexample does not seem to exist: there is no way to assign numbers between 1 and 60 to 9 people such that there would not be two subsets with the same sum.  In fact, there doesn't seem to any 8-person counterexample either.
7-person counterexamples are easy to find, though: $(1, 2, 4, 24, 40, 48, 56)$ and $(60, 59, 58, 56, 53, 47, 36)$ are two of them.  So now the interesting question becomes, is there some way to prove that an 8-person counterexample cannot exist without an exhaustive search?
A: A group of $9$ is sufficient to get matching age groups, using a small extension on estimating the range of possible sums.
As observed the number of non-empty subsets will be $2^9-1=511$ and as before we can assume that all ages are different, since otherwise we can just use two people with the same age to achieve our goal. 
Now consider the youngest person, with an age of $y$ - this is the lower limit of possible age sums. The upper limit is then at most $y+\sum_{53}^{60}i = y+452$. There are thus $453$ or fewer different age sums available for different groups and thus the pigeonhole principle applies, where two groups must have the same age sum.

Clearly we can extend the age limit also: $9$ people of up to age $67$ will still produce groups with a common age sum from this still fairly simple process. 
At the cost of sacrificing a few less-useful subsets, we can reduce the totals range further also. The full set 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 example for $10$ people, considering the $1013$ groups defined under this process we can impose an age limit of $130$ 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.
The corresponding age limits for some smaller groups are:
$$\begin{array}{c|c|c}
\text{# people} & \text{# useful subsets} & \text{age limit}\\ \hline
10 & 1013 & 130 \\
9 & 502 & 75\\
8 & 247 & 44\\
7 & 120 & 26\\
6 & 57 & 16\\
5 & 26 & 10
\end{array}$$
But note that these age limits are likely underestimates. For example, in the last case, the logic to achieve this has assumed we have ages of $\{10,9,8,7\}$ which immediately produce two groups of equal age sum ($10+7=9+8$). In fact leaving only one "young end" age can produce huge gaps in the possible age sums (fewer holes for our pigeons). So it is entirely likely that the age limits are somewhat higher and I am not surprised by the claim that a group of $8$ could potentially have an age limit of $60$ and still necessarily have same-age-total groups.
A: In your solution, you have mentioned that the maximum sum is 60. But notice that if there were two people of the same age then we are done as we can easily choose the two persons as a different group. So it should be distinct, i.e., max sum= $\displaystyle\sum_{i=0}^9 (51+i)=\dfrac{10}{2}(51+60)=555$. But here comes the surprise, the minimum sum = $\displaystyle\sum_{i=0}^9 (1+i)=\dfrac{10}{2}(1+10)=55$, and the total number of sums using inclusion and exclusion principle we get, $555-55+1=501$. Here see that $2^9-1=511$, hence  @Ilmari Karonen did not encounter any problem with 9 number set with his brute force algorithm.
But we are not done here, notice that $2^8-1=255<501$. It simply implies that we can always find a subset such that no 2 groups of people will have the same age. Here we have reduced the number of possible sums, which is very important as no we get the reduced possibilities ignoring all that impossible cases. You can try to prove that 8- number set is not possible to obtain.
