Solution:
There is no platinum matrix for $n=1$ and for $n=2$. We claim that a platinum matrix exists for all integers $n \geq 3$.

Define a transversal as a collection of $n$ matrix entries which are taken from distinct rows and columns, and which contains each of the numbers $1$ to $n$.

When $n \geq 3$ is odd, we can construct an $n \times n$ platinum matrix in the following manner: First, fill up the first row with $1, n, n-1, \ldots, 3, 2$ from left to right. Then fill up each "diagonal" (which goes up to down, left to right and wraps back to the 1st column after the $n$th column) by starting with its first row entry and incrementing by $1$ as we go down to the last row. From this point onwards, entries are reduced $\bmod n$, but with $n$ written instead of $0$. For example, for $n=3$, we have the following platinum matrix.
$$\left(\begin{array}{lll} 1 & 3 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{array}\right)=\left(\begin{array}{lll} 1 & & \\ & 2 & \\ & & 3 \end{array}\right)+\left(\begin{array}{lll} 3 & \\ & & 1 \\ 2 & & \end{array}\right)+\left(\begin{array}{ll} & \\ 3 & \\ & 1 \end{array}\right)$$
From the construction described, the entry in row $i$, column $j$ ($1 \leq i, j \leq n$) is $a_{ij}=2i-j$ (again, reduced $\bmod n$ as mentioned above). For a fixed row $i$, the entries for different columns $j$ and $j'$ are distinct. For a fixed column $j$, the entries for different rows $i$ and $i'$ are distinct since $n$ is odd. The diagonal whose first row entry is $2$ is a transversal; in fact, each diagonal is a transversal. Thus, the matrix is platinum.

Let $n \geq 8$ be even, so $n-3 \geq 5$ is odd. Consider the $(n-3) \times (n-3)$ platinum matrix, denoted by $C_{n-3}$, following the construction above. We start constructing our $n \times n$ platinum matrix as follows:
$$\left(\begin{array}{c|c} C_{n-3} & \\ \hline & C_3 \end{array}\right) \quad \text{where} \quad C_3=\left(\begin{array}{ccc} n-2 & n & n-1 \\ n & n-1 & n-2 \\ n-1 & n-2 & n \end{array}\right)$$
We then need to fill in the $3 \times (n-3)$ matrix and the $(n-3) \times 3$ matrix adjacent to $C_{n-3}$. To do this, from the $(n-3)-1=n-4 \geq 4$ diagonals of $C_{n-3}$ other than its main diagonal, choose $3$. For one of these transversals, project its entries vertically into an empty row and horizontally into an empty column, then replace all of the entries of this chosen transversal by $n-2$. Then repeat this procedure using the symbols $n-1$ and $n$ using the two other transversals. The resulting matrix is then a platinum matrix. Of the $n-4 \geq 4$ transversals described above for $C_{n-3}$, at least one has not been used yet. This, along with a "transversal" of $C_3$ other than its main diagonal, then form a transversal of the formed $n \times n$ matrix; none of the entries of this transversal are in the main diagonal. The matrix formed is thus platinum.

The following illustrates the construction for the case $n=8$. The final matrix is platinum, with the boxed entries all off-diagonal and forming a transversal.

Lastly, consider the following matrices.
$$\left(\begin{array}{llll} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1 \\ 2 & 1 & 4 & 3 \end{array}\right) \quad \left(\begin{array}{llllll} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 3 & 6 & 1 & 4 & 2 \\ 4 & 1 & 5 & 2 & 6 & 3 \\ 2 & 4 & 1 & 6 & 3 & 5 \\ 6 & 5 & 4 & 3 & 2 & 1 \\ 3 & 6 & 2 & 5 & 1 & 4 \end{array}\right)$$
It is straightforward to verify that these are platinum matrices for $n=4$ and $n=6$.