1. Let $a_{1}, a_{2}, \ldots, a_{9}$ be nine real numbers, not necessarily distinct, with average $m$. Let $A$ denote the number of triples $1 \leq i<j<k \leq 9$ for which $a_{i}+a_{j}+a_{k} \geq 3 \mathrm{~m}$. What is the minimum possible value of $A$ ?
Proposed by Ray Li.
Answer. $A \geq 28$.
Solution 1. Call a 3-set good iff it has average at least $m$, and let $S$ be the family of good sets.
The equality case $A=28$ can be achieved when $a_{1}=\cdots=a_{8}=0$ and $a_{9}=1$. Here $m=\frac{1}{9}$, and the good sets are precisely those containing $a_{9}$. This gives a total of $\binom{8}{2}=28$.
To prove the lower bound, suppose we have exactly $N$ good 3 -sets, and let $p=\frac{N}{\binom{9}{3}}$ denote the probability that a randomly chosen 3 -set is good. Now, consider a random permutation $\pi$ of $\{1,2, \ldots, 9\}$. Then the corresponding partition $\bigcup_{i=0}^{2}\{\pi(3 i+1), \pi(3 i+2), \pi(3 i+3)\}$ has at least 1 good 3 -set, so by the linearity of expectation,
$$
\begin{aligned}
1 & \leq \mathbb{E}\left[\sum_{i=0}^{2}[\{\pi(3 i+1), \pi(3 i+2), \pi(3 i+3)\} \in S]\right] \\
& =\sum_{i=0}^{2}[\mathbb{E}[\{\pi(3 i+1), \pi(3 i+2), \pi(3 i+3)\} \in S]] \\
& =\sum_{i=0}^{2} 1 \cdot p=3 p .
\end{aligned}
$$

Hence $N=p\binom{9}{3} \geq \frac{1}{3}\binom{9}{3}=28$, establishing the lower bound.
This problem and solution were proposed by Ray Li.
Remark. One can use double-counting rather than expectation to prove $N \geq 28$. In any case, this method generalizes effortlessly to larger numbers.
Solution 2. Proceed as above to get an upper bound of 28 .
On the other hand, we will show that we can partition the $\binom{9}{3}=843$-sets into 28 groups of 3 , such that in any group, the elements $a_{1}, a_{2}, \cdots, a_{9}$ all appear. This will imply the conclusion, since if $A<28$, then there are at least 57 sets with average at most $m$, but by pigeonhole three of them must be in such a group, which is clearly impossible.
Consider a 3 -set and the following array:
\begin{tabular}{lll}
$a_{1}$ & $a_{2}$ & $a_{3}$ \\
$a_{4}$ & $a_{5}$ & $a_{6}$ \\
$a_{7}$ & $a_{8}$ & $a_{9}$
\end{tabular}

Consider a set $|S|=3$. We obtain the other two 3-sets in the group as follows:
- If $S$ contains one element in each column, then shift the elements down cyclically mod 3 .
- If $S$ contains one element in each row, then shift the elements right cyclically mod 3 . Note that the result coincides with the previous case if both conditions are satisfied.
- Otherwise, the elements of $S$ are "constrained" in a $2 \times 2$ box, possibly shifted diagonally. In this case, we get an L-tromino. Then shift diagonally in the direction the L-tromino points in.

One can verify that this algorithm creates such a partition, so we conclude that $A \geq 28$.
This second solution was suggested by Lewis Chen.