The only problem I solved, hooray! Basically, you just show that at least one out of three triples belong to A, because otherwise we arrive at a contradiction. Then you just construct a set for which this is true, for example: $1,1,1,1,1,1,1,1,2$ EDIT: 300th post!

infiniteturtle's "one out of three" made it trivial :/ Note that the sum of all $a_i$ is $9m$. By the real number version of Pigeonhole Principle, if three numbers $x,y,z$ add up to $9m$, one of them must be at least $3m$. In particular, if we partition the $a_i$ into three triples and let $x,y,z$ be the sum of the elements in the first, second, triple respectively, at least one of these three triples sum to at least $3m$. So at least 1/3 of all triples have their sum at least $3m$. The number of triples is obviously $\binom{9}{3} = 84$, so at least $28$ triples have their sum at least $3m$ and hence $A \ge 28$. It remains to show that this is achievable; see infiniteturtle's construction. If a triple contains all 1s, its sum is $3 < \dfrac{10}{3} = 3 \cdot \dfrac{10}{9} = 3m$. There are $\binom{8}{3} = 56$ triples containing solely 1s, so $A \le 84-56 = 28$. EDIT: Sniped

This is more-or-less borntobeweild's solution. Its what I would have turned in had I competed. I claim the answer is $28$. A construction is arrived at by considering $a_1 = a_2 = ... = a_8 = -1$ and $a_9 = 8$. Consider a random permutation $\sigma \in S_9$. Now consider the triplet $\{a_{\sigma(1)}, a_{\sigma(2)}, a_{\sigma(3)}\}$. It is clear that as least one of $\{a_{\sigma(1)}, a_{\sigma(2)}, a_{\sigma(3)}\}, \{a_{\sigma(4)}, a_{\sigma(4)}, a_{\sigma(6)}\}, \{a_{\sigma(7)}, a_{\sigma(8)}, a_{\sigma(9)}\}$ has a sum of elements at least $3m$ since the sums of all three sums equals $9m$. It follows that $\text{Pr}[a_{\sigma(1)} + a_{\sigma(2)} + a_{\sigma(3)} \ge 3m] \ge \frac{1}{3}$. It follows that the number of triplets whose sum is at least $3m$ numbers at least \[ \frac{1}{3} \dbinom{9}{3} = 28\] as desired.

I get it, but how are we sure that it is possible to partition the 84 triples into 28 sets such that in each set, the numbers a_1, a_2, ..., a_9

The solutions of borntobeweild and dinoboy do not require such a partition. chaotic_iak's solution does not explicitly require such a partition, but it is rather muddy in the relevant spot and probably would be docked points on a competition for this reason.

There is actually a fairly nice way to explicitly construct a partition based on looking at shapes in a $3 \times 3$ grid, but it doesn't generalize easily, unlike the probabilistic / double-counting solutions.

Here's how I thought about it before I dropped out. I knew that two of the three numbers, at best, had to be greater than the average. That's really all I got. I had no idea how to proceed.

darn I had it, but I thought the solution was too simple so I must have gotten it wrong and didn't bother to turn it in. Anyway, here's what I did (it's probably unnecessarily complicated).

"obviously" is not that obvious (and in fact one of the main points of the proof). It also suffers from the same mistake in my proof (prove that each triple indeed appears $\binom{6}{3}$ times). SZRoberson is even more wrong. The answer is 1/3 of the number of triples, not 2/3.

My formal solution which I submitted was almost exactly like exmath89's (he worded it better than I did) except that I divided by $3!$ and $10$ at the same time.

The answer is $28$.To prove that the answer is,just observe there are $84$ triples and if we choose $3$ triplets $ai,aj,ak$ $ah,ag,as$ $ar,at,am$ such that $i,j,k,h,g,s,t,m$ such that all are pairwaise distinct,at least one will we have sum at least $3m$,so from this we easy get the answer is $28$(at least one third of all triples will have the sum at least $3m$ ).An example is easy,just take the biggest number be much larger than the others(for examle,$4$ and $9$ zeroes).

I misread the question but realized that what I had misunderstood was probably a question itself Can we maximize the value of $A$ ?

Well that is obvious,took all numbers be equal,than you have all triplets,or $84$.

It's just Baranyai's theorem: If $k$ divides $n$, the set of all $\binom{n}{k}$ $k-$subsets of an $n-$set may be partitioned into disjoint parallel classes $A_{i}$ for $i=1, 2,\dots , \binom{n-1}{k-1}$.

bump....

Note that because the average value of the values $a_n$ is $m$, you could say that the expected value of $a_n$ is $m$. If you were to take three disjoint triples, note that as the expected value of the sum of each triple is $3m$, you would expect that at least one of the triples has a sum that is greater than or equal to $3m$. From this, we can infer that at least $\frac{1}{3}$ of the possible triples satisfy $a_i+a_j+a_k \geq 3m$, which implies $A \geq 28$ as there are $\binom{9}{3} = 84 = 28 \cdot 3$ triples. A possible construction is if the nine real numbers are $8$ instances of the number $1$ and $1$ instance of the number $2$. Note that in order for a triple to satisfy the condition, it must have the $2$ in it, and there are $\binom{8}{2} = 28$ ways to choose the remaining numbers. So, the minimum possible value for $A$ is $28$.

The answer is $28$. A construction is $a_1=a_2=\cdots=a_8=0$ and $a_9=187$, in which case a triple $1 \leq i<j<k\leq 9$ satisfies $a_i+a_j+a_k \geq 3m$ if and only if $k=9$. We can easily check that there are $\binom{8}{2}=28$ such triples. Now I will show that $A$ cannot be smaller. Instead of viewing $i,j,k$ as triples, view them as subsets of $\{1,2,\ldots,9\}$. We can partition all of the possible sets into groups of 3 sets $A,B,C$ such that $A \cap B=\emptyset,B\cap C=\emptyset, C \cap A=\emptyset$: an example is $A=\{1,5,6\},B=\{2,8,9\},C=\{3,4,7\}$. Since $A \cup B \cup C$ is then always $\{1,2,\ldots,9\}$, we require at least one of the sets to result in a sum of at least $3m$, otherwise their combined sum is less than $9m=1+2+\cdots+9$: impossible. Thus we have at least $\frac{1}{3}\binom{9}{3}=28$ triples, as desired. $\blacksquare$

Ok so at least one out of every three disjoint triples must have a sum at least $3m$ This gives $A \geq \frac{1}{3}\binom{9}{3}=24$. now we just find a construction.

We claim that the answer is 28. We say that a triple is acceptable if its average is greater than or equal to the overall average. For a construction of 28, take $$(10^{100},1,1,1,1,1,1,1,1).$$Clearly, a triple is acceptable if and only if it contains $10^{100}$, so there are 28 acceptable triples. Consider partitioning the 9 numbers into 3 sets of 3. Note that there is at least 1 acceptable set out of these 3 due to Pidgeonhole. Therefore, the expected number of acceptable sets when doing a random partition is at least 1. By Linearity of Expectation, the expected number of acceptable sets when doing a random partition is $$n/28,$$where $n$ is the number of total acceptable triples, since each acceptable triple has a 1/28 chance of appearing in the set. Therefore, $n\geq28$, so we are done.

Call a triple fat if it satisfies the desired condition. The answer is $28$ triples. For a construction, just make one of the $a_n$ really big. Note that among any three distinct triples, at least one is fat. Thus, among all $\frac{9!}{(3!)^4}$ partitions of $[9]$ into triples, each triple appears in exactly $\frac 12{6 \choose 3} = 10$ partitions. Thus, to contribute $280$ fat triples among partitions with multiplicity, there must exist at least $28$ triples, as needed.

The answer is $28$, achievable by taking $a_1=a_2=\dots=a_8=0$ and $a_9=9$. We will show that we must have $A \ge 28$. Note that there are $\tbinom{9}{3}=84$ possible triples $(i, j, k)$ possible. Choose a permutation $\sigma$ of $(a_1, \dots, a_9)$ randomly. Let $S_i=\{a_{\sigma(3i+1)}, a_{\sigma(3i+2)}, a_{\sigma(3i+3)}\}$ for $i=0, 1, 2$. Let $X$ be the number of $S_0, S_1, S_2$ that have the sum of its elements at least $3m$. Clearly $X \ge 1$, so $\mathbb{E}[X] \ge 1$, and if the probability of a randomly selected triple having sum at least $3m$ is $p$, then $\mathbb{E}[X] = 3p$. Thus $3p \ge 1$, so $p \ge \tfrac{1}{3}$, which means that $A = 84p \ge 84 \cdot \frac{1}{3} = 28$, as needed.

Did anyone manage to create 28 triples of disjoint (i, j, k)?

The answer is $\boxed{28}$, achieved by $a_1=a_2=\dots=a_8=0$ and $a_9=1434$. Among any three disjoint triples, at least one satisfies the condition (else $\sum a_i < 9m$, contradiction). There are $280$ triples of disjoint triples so thus there are at least $280$ triples satisfying the condiiton. But every triple is counted in $\frac12 \binom63 = 10$ sets, so we have at least $28$ triples, as desired.

The answer is 28, achieved by taking $(a_1,a_2,\dots ,a_8,a_9)=(0,0,\dots ,0,1)$. To see the bound, notice that amongst any 3 disjoint triples, atleast one of them satisfies the desired property. Its not so hard to see that there are $280$ disjoint triples so there are atleast 280 triples satisfying the given condition counted with multiplicity, and since each triple is counted $10$ times, we get that there are atleast $28$ triples as desired.