, 2016

The minimal $n$ in question is $n=9$.
First we show that each number $k \in \{1, 2, ..., n\}$ occurs in the table at most $34$ times. Let a row contain at least two $k$'s. Underline all of them except the rightmost one. The remaining numbers in the table are not underlined. By hypothesis $k$ does not occur above an underlined number. It follows that the underlined numbers are in different columns, hence there are at most $17$ of them. In addition, by construction each row without underlined $k$'s contains at most one $k$, so there are also at most $17$ numbers $k$ that are not underlined. In summary the table has at most $17+17=34$ numbers $k$, as stated.

Since each $k \in \{1, 2, ..., n\}$ occurs in the table $x_k \le 34$ times, the equality $x_1 + \ldots + x_n = 17^2$ yields $n \ge \frac{17^2}{34}$, meaning that $n \ge 9$.

For an example with $n=9$ consider the diagonals parallel to the main diagonal containing the bottom left and the top right cell. Label them consecutively $1$, $2$, ..., $33$ so that the top left cell is diagonal $1$ and the bottom right cell is diagonal $33$. For each $k=1, 2, ..., 9$ write $k$ in all cells of diagonals $2k-1$ and $2k$; for each $k=1, 2, ..., 7$ write $k$ in all cells of diagonals $2k+1$ and $2k+18$; finally write $8$ in the only cell of diagonal $33$. In every row and column there are at most two numbers equal to a given $k \in \{1, 2, ..., 9\}$; and if there are two of them then they are adjacent. It follows that the table satisfies the given conditions.