To find the number of unique sums, we first need to find how many different ways there are to choose three numbers from the set. This is just $\binom{7}{3}=35$. Thus, we just need to find how many different sums we can get when choosing three numbers.
Ordering the numbers in ascending order, we get $4,6,8,10,12,14,16$, and notice that each number must be used so that each sum will be unique. To see this, note that we can get an upper bound by taking the sum of three consecutive numbers, which would be $4+5+6=15$, and our smallest number is greater than this. Thus, the sums must be greater than $15$.
Starting with $4$, we can get $4+5=9$. We can get all other positive integers by adding $4$ to one of the following: $4+5, 4+5+7, 4+5+7+9, 4+5+7+9+11, 4+5+7+9+11+13$.
This gives $9,11,12,13,14,16,17,18,19,20,21,22,23,24,25,26,27,28$, $29,30,31,32,33,34,35$. These are $\${13}$ distinct sums.
Since the range is $21-9+1=13$, some numbers were found multiple times. For example, $24=4+10+10$ can be found in two ways using $4$, and $11=4+5+2$ can be found in 4 ways using $4$. This gives $\boxed{28}$ as the answer.
How many unique sums can be formed by adding any three different numbers from the set $\{4,6,8,10,12,14,16\}\,?$
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