3. Given $n(n>2)$ coins of different weights and $n$ scales. Each time during weighing, one can choose a scale, place one coin on each end, compare their weights, and then remove the coins. It is known that one of the scales is faulty (but it is unknown which one), and it sometimes gives the correct result and sometimes gives the wrong result. Question: What is the minimum number of weighings required to definitely find the heaviest coin?

---

Note: The translation preserves the original text's line breaks and format.
3. At least $2n-1$ weighings are required.

First, we prove that after $2n-1$ weighings, the heaviest coin can definitely be found. More precisely, by induction on $n$, we prove that if there are three scales, one of which may be faulty, the heaviest coin can be determined through $2n-1$ weighings from $n \geq 2$ coins.

When $n=2$, place the two coins on the ends of the three scales sequentially. If a balance occurs in any weighing, then that scale is faulty, and the heavier coin can be determined from any of the other two weighings; if no balance occurs, then one of the coins must be shown to be heavier twice, which is the heavier coin, because an incorrect result can only occur once. This is the base case for induction.
Now assume $n \geq 3$. Pick two coins and two scales.
First, use these two scales to compare the two coins. Place the two coins consistently on the left and right ends of the two scales (both times on the same end). There are two possible scenarios.
(1) On both scales, the same coin is heavier, denoted as $a$, and the other coin as $b$. Since at least one of the two scales is good, $a$ is indeed heavier than $b$. This indicates that $b$ cannot be the heaviest. Thus, the problem reduces to finding the heaviest coin from $a$ and the $n-2$ coins that did not participate in the first two weighings. By the induction hypothesis, for $n-1$ coins, no more than $2n-3$ weighings are needed. Including the initial two, a total of no more than $2n-1$ weighings are required.
(2) On the two scales, one is balanced or the two scales give contradictory results: one shows one coin is heavier, and the other shows the other coin is heavier. These results indicate that at least one of the two scales is faulty. At this point, take the third scale, which is definitely good. Using it, the heaviest coin can be found through $n-1$ weighings: first compare the two initial coins, take the heavier one and compare it with the third coin, then take the heavier one and compare it with the fourth coin, and so on until the last. Including the initial two, a total of $n+1$ weighings are performed.

Next, we prove that if fewer than $2n-1$ weighings are performed, the heaviest coin cannot be guaranteed to be found. To do this, we only need to prove that $2n-2$ weighings are not sufficient to determine the heaviest coin, because an additional weighing can be added arbitrarily and its result ignored. Assume otherwise, that there exists an operation that can determine the heaviest coin through $2n-2$ weighings.

Number the coins as $1, 2, \cdots, n$. Perform the first $2n-3$ weighings in the operation. Assume each weighing shows the coin with the larger number is heavier. By the pigeonhole principle, among the $2n-3$ weighings, one of the coins numbered $1, 2, \cdots, n-1$ is eliminated at most once (each time one lighter coin is eliminated), let its number be $k$.
Below, we provide a method to produce such a result.
In fact, at least the following two scenarios can lead to this result.
(i) The coins are placed on the scales in increasing order of weight, and all scales (including the faulty one) show the correct result.
(ii) Except for the coin numbered $k$ (which is the heaviest), all other coins are placed on the scales in increasing order of weight. The scale that eliminates the coin numbered $k$ is faulty because it shows an incorrect result, while all other weighings show the correct result.

Now consider whether the coin numbered $k$ participates in the final $2n-2$th weighing.
The coin numbered $k$ does not participate in the final $2n-2$th weighing, still assuming each weighing shows the coin with the larger number is heavier. In this case, scenarios (i) and (ii) are still possible.
The coin numbered $k$ participates in the final $2n-2$th weighing, assuming it shows as heavier. In this case, on the one hand, it could be scenario (i), and the last weighing is performed on the faulty scale. On the other hand, it could be scenario (ii), and the last weighing is performed on a good scale.

In this way, regardless of the result of the remaining weighing, scenarios (i) and (ii) are still possible. Thus, both the coin numbered $k$ and the coin numbered $n$ could be the heaviest.