The multi-legged beings' international conference is to be held at the top of Glass Mountain. The participants have $a_{1}, \ldots, a_{n}$ legs respectively, all positive even numbers. Glass Mountain is slippery, and the beings gathered at the foot of the mountain can only climb up if they put on special climbing shoes on at least half of their legs. Knowing that climbing shoes can only be brought up the mountain and back down if worn on legs, how many shoes are at least needed for the conference participants to reach the top?
Solution. The numbering of the conference participants can be freely chosen, so without loss of generality, we can assume that the number of legs is ordered as $a_{1} \leq a_{2} \leq \ldots \leq a_{n}$, where the numbers $a_{i}\left(i=1,2, \ldots, n \in \mathbb{N}^{+}\right)$ are positive even numbers.

It is obvious that if the multi-legged beings can climb to the top of the mountain with a certain number of shoes in some order, they can also come down with the same number of shoes in reverse order. Therefore, it is sufficient to only examine how many shoes are minimally required for the participants to reach the top.

For $n=1$, based on the problem's conditions, the single participant needs at least $\frac{a_{1}}{2}$ shoes to climb the mountain.

For $n \geq 2\left(n \in \mathbb{N}^{+}\right)$, we will show that the participants need

$$
\max \left\{\frac{a_{1}+a_{2}}{2}, \frac{a_{n}}{2}\right\}
$$

shoes to attend the conference.

In the first round, we send the 1st and 2nd participants to the top of the mountain with $\frac{a_{1}}{2}$ and $\frac{a_{2}}{2}$ shoes on their feet, respectively, and then the 2nd participant, putting all the shoes that have reached the top on their feet, brings them down. This is feasible based on the estimate $\frac{a_{1}+a_{2}}{2} \leq \frac{2 a_{2}}{2}=a_{2}$. Thus, all the shoes will be at the foot of the mountain.

Next, the $n$-th multi-legged participant goes to the top with $\frac{a_{n}}{2}$ shoes. Since

$$
\frac{a_{n}}{2} \leq \max \left\{\frac{a_{1}+a_{2}}{2}, \frac{a_{n}}{2}\right\}
$$

this is also feasible. Then, the 1st multi-legged participant, who is at the top, walks back and forth to bring all the shoes to the foot of the mountain, putting $a_{1}$ shoes on their feet when going down and $\frac{a_{1}}{2}$ shoes on their feet when going up. If on the last descent, they do not have $a_{1}$ shoes on their feet, they should come down with whatever is left. This will definitely be more than $\frac{a_{1}}{2}$, since they already carried this amount on their feet when going up.

Then, the procedure described above continues with the $(n-1)$-th multi-legged participant. They go up together with the 1st and 2nd participants with $\frac{a_{1}}{2}$ and $\frac{a_{2}}{2}$ shoes, respectively, and then the 2nd participant comes down with $\frac{a_{1}+a_{2}}{2}$ shoes on their feet. Thus, all the shoes are brought back to the foot of the mountain, and then the $(n-1)$-th multi-legged participant goes to the top with $\frac{a_{n-1}}{2}$ shoes. This is feasible because

$$
\frac{a_{n-1}}{2} \leq \frac{a_{n}}{2} \leq \max \left\{\frac{a_{1}+a_{2}}{2}, \frac{a_{n}}{2}\right\}
$$

Then, the 1st participant, using the previously mentioned method, shuttles back and forth to bring all the shoes to the foot of the mountain.

Continuing this algorithm, the $(n-2)$-th, $(n-3)$-rd, and finally the 2nd and 1st multi-legged participants will permanently reach the top of the mountain. As mentioned earlier, the descent happens in reverse order. This proves that the given number of shoes ensures the participants' ascent. Next, we will show that fewer shoes cannot ensure the participation of the multi-legged beings in the conference.

Since the $n$-th participant needs at least $\frac{a_{n}}{2}$ shoes to reach the top of the mountain, the number of shoes must be at least

$$
\frac{a_{n}}{2}
$$

Furthermore, for at least two participants, if the first two to reach the top are the $i$-th and $j$-th participants $\left(1 \leq i<j \leq n, \quad i, j \in \mathbb{N}^{+}\right)$, then at least $\frac{a_{i}}{2}+\frac{a_{j}}{2}$ shoes are needed for their ascent, since either they go up together or separately, but in the latter case, the second participant cannot bring shoes from the top. If the first participant returned from the top, they could leave at most one shoe, but this would not reduce the number of shoes brought to the point where at least two participants are at the conference site. Since

$$
\frac{a_{i}}{2}+\frac{a_{j}}{2} \geq \frac{a_{1}}{2}+\frac{a_{2}}{2}
$$

considering condition (1), we have shown that the minimum number of shoes required cannot be less than $\max \left\{\frac{a_{1}+a_{2}}{2}, \frac{a_{n}}{2}\right\}$.