Consider the matrix
$$
\left(a_{i j}\right)_{n \times n}\left(a_{i j} \in\{1,2,3\}\right) \text {. }
$$

If $a_{i j}$ is such that its row $i$ and column $j$ both contain at least three elements (including $a_{i j}$) that are equal to $a_{i j}$, then the element $a_{i j}$ is called "good". If the matrix $\left(a_{i j}\right)_{n \times n}$ contains at least one good element, find the minimum value of $n$.
The minimum value of $n$ is 7.
When $n=6$, the matrix
$\left(\begin{array}{llllll}1 & 1 & 2 & 2 & 3 & 3 \\ 1 & 1 & 3 & 3 & 2 & 2 \\ 2 & 2 & 1 & 1 & 3 & 3 \\ 2 & 2 & 3 & 3 & 1 & 1 \\ 3 & 3 & 2 & 2 & 1 & 1 \\ 3 & 3 & 1 & 1 & 2 & 2\end{array}\right)$
has no good elements, so $n \geqslant 7$.
Below, we use proof by contradiction to show that when $n=7$, the matrix $\left(a_{i j}\right)$ must have at least one good element.

Assume that no element in the matrix $\left(a_{i j}\right)$ is good.
Define: If a number $m$ appears at least three times in a row (or column), then we say that the number $m$ controls that row (or column).
It is easy to see that the number of rows (or columns) controlled by the number $m$ is at most 4.
First, we prove a lemma:
Lemma: If the number 1 controls the first three rows, then the 1s in these three rows must appear in at least five columns.

Proof: Since the number 1 appears at least 9 times in the first three rows, by the pigeonhole principle, it must appear in at least five columns. Now, we show that five columns are impossible.
If the number 1 appears in the first five columns, consider the submatrix $T$ formed by the first five columns and the last four rows of the original matrix. Then, $T$ contains at most one 1.
If there is a row in $T$ where 2 appears at least three times and another row where 3 appears at least three times, then there must be a column where 1, 2, and 3 each appear three times. Therefore, in this column, 1, 2, and 3 each appear at most twice, which is a contradiction.
Assume that there are three rows in $T$ where 2 appears at least three times, then in the first five columns, the numbers 1 and 2 each appear at most twice in each column. Thus, in the first five columns, the number 3 appears at least three times in each column, which is a contradiction.
Therefore, this scenario is impossible.
We prove in two steps:
(1) If the number $m$ controls at least 3 rows (or columns), then the number of elements in the matrix $\left(a_{i j}\right)$ that are equal to $m$ is at most 16.
We discuss two cases (assume $m=1$ without loss of generality).
(i) The number 1 controls four rows.
Assume these are the first four rows, then the number 1 appears at least 12 times in the first four rows. By the pigeonhole principle, it must appear in at least six columns.
(1) If the number 1 appears in seven columns, then the number of 1s is at most 14.
(2) If the number 1 appears in six columns, then the number of 1s is at most $6 \times 2 + 3 = 15$.
(ii) The number 1 controls three rows.
Assume these are the first three rows. By the lemma, the number 1 appears in at least six columns.
(1) If the number 1 appears in seven columns, then the number of 1s is at most 14.
(2) If the number 1 appears in six columns, then the number of 1s is at most $6 \times 2 + 4 = 16$.
(2) For $m=1,2,3$, the number of rows or columns controlled by $m$ is at least 3.
By the pigeonhole principle, assume 1 controls at least 3 rows.
We show that the number of rows or columns controlled by 2 and 3 is at least 3.
We discuss two cases:
(i) The number 1 controls four rows.
Assume these are the first four rows, then the number 1 appears at least 12 times in the first four rows. By the pigeonhole principle, it must appear in at least six columns.
(1) If the number 1 appears in seven columns, then 2 and 3 each control three columns and four columns, respectively.
(2) If the number 1 appears in the first six columns, consider the submatrix $T$ formed by the first six columns and the last three rows of the original matrix. $T$ contains only 2 and 3.
If the first row of $T$ contains 3 twos and 3 threes, then 2 and 3 each control three columns; if the first row of $T$ contains 4 twos, then 3 controls four columns, and 2 controls three rows (since in the last two rows of $T$, it is impossible for one row to contain more than 3 threes).
(ii) The number 1 controls three rows.
Assume these are the first three rows. By the lemma, the number 1 appears in at least six columns.
(1) If the number 1 appears in seven columns, then 2 and 3 each control three columns and four columns, respectively.
(2) If the number 1 appears in the first six columns, consider the submatrix $T$ formed by the first six columns and the last four rows of the original matrix. $T$ contains at most 3 ones, so there must be a row in $T$ that does not contain 1.
If this row contains 3 twos and 3 threes, then 2 and 3 each control three columns; if this row contains 4 twos, then 3 controls four columns.
Since 1 appears at most 3 times, 2 must control at least three rows in the last four rows.
By (1) and (2), in the matrix $\left(a_{i j}\right)$, the number of rows or columns controlled by the numbers $m=1,2,3$ is at least 3.
Thus, the numbers 1, 2, and 3 each appear at most 16 times in the matrix $\left(a_{i j}\right)$.
But when $n=7$, the matrix $\left(a_{i j}\right)$ has 49 elements, which is a contradiction.
Therefore, when $n=7$, the matrix $\left(a_{i j}\right)$ must have at least one good element.