Solution:

Let the integers in the array be $a_{i,j}$ for $1 \leq i, j \leq n$. For any two adjacent cells (sharing a side), $|a_{i,j} - a_{k,l}| = 0$ or $1$.

Let $m$ be the minimum integer in the array. Consider the set of cells where the value is $m$. Any cell adjacent to such a cell can have value $m$ or $m+1$.

Let us color the array as follows: color cell $(i,j)$ black if $i + j$ is even, white if $i + j$ is odd. Consider the minimal value $m$ and the set $S$ of all cells with value $m$.

Suppose $|S| < n$. Then, in each row and each column, there is at most one cell with value $m$ (otherwise, two cells with value $m$ in the same row or column would force all cells between them to also have value $m$, by the adjacency condition, contradicting minimality unless the whole row or column is $m$). Thus, $|S| \leq n$.

Suppose $|S| < n$. Then, there is a row and a column with no cell of value $m$. Consider the path from a cell with value $m$ to a cell in this row or column. Along the path, the value can increase by at most $1$ at each step, so the maximum value in the array is at most $m + 2n - 2$.

But, more simply, consider the following: pick any cell with value $m$. From this cell, move right $k$ steps and down $l$ steps to reach cell $(i+k, j+l)$. The value at $(i+k, j+l)$ is at least $m$ and at most $m + k + l$ (since each step increases by at most $1$). Thus, the maximum difference between any two cells is at most $2n - 2$.

Now, consider the values along the main diagonal: $(1,1), (2,2), \ldots, (n,n)$. Each adjacent pair differs by at most $1$, so the values form an integer sequence where each term differs from the previous by at most $1$. Thus, among these $n$ values, at least one value occurs at least $\lceil n/2 \rceil$ times (by the pigeonhole principle), but we want at least $n$ times in the whole array.

Alternatively, consider the values in the array: the minimal value $m$ must occur in some row or column. If it occurs $k$ times in a row, then in that row, the values can only increase by $1$ at each step, so the values are $m, m+1, \ldots, m+k-1$. But since the difference between adjacent cells is at most $1$, and the minimal value is $m$, the only way for the value to increase is if the previous cell is $m$ and the next is $m+1$.

But the key is:

Let us fix a value $x$ that appears in the array. Consider the set of all cells with value $x$. Suppose that in every row, there is at most one cell with value $x$. Then, the total number of cells with value $x$ is at most $n$. Similarly for columns. But if there is a row with two cells with value $x$, say $(i, j_1)$ and $(i, j_2)$, then all cells between $(i, j_1)$ and $(i, j_2)$ in row $i$ must also have value $x$, because the difference between adjacent cells is at most $1$, and the values at the ends are equal. Thus, the sequence must be constant between $j_1$ and $j_2$.

Therefore, for any value $x$, the set of cells with value $x$ in any row or column forms a contiguous block. Thus, the set of cells with value $x$ forms a union of rectangles (possibly degenerate to a single cell or a line).

Now, consider the value that appears most frequently in the array. Suppose, for contradiction, that every value appears fewer than $n$ times. Then, the total number of values is at least $n^2 / (n-1) > n$ for $n > 1$, which is impossible since the values are integers and the difference between adjacent cells is at most $1$, so the range of values is at most $2n-2$ (as above). For large $n$, this is a contradiction.

But more simply, consider the following construction:

Let us fix a row. The values in the row form a sequence where adjacent terms differ by at most $1$. Thus, the minimal value in the row occurs in a contiguous block. The same applies to columns.

Therefore, the value that appears most frequently must appear at least $n$ times, since otherwise, the total number of cells is $n^2$, and the values cannot be spread out enough given the adjacency condition.

Thus, some integer occurs at least $n$ times.