It is always possible, we can place the $\binom{n}{2}$ pairs in a $n \times n$ square when $n$ is odd and in a $(n-1) \times n$ rectangle when $n$ is even.
This problem is equivalent to the edge coloring problem for complete graph $K_n$. Look at wiki for the geometric intuition underlying following construction.
Let $[n]$ be a short hand for $\{ 0, \ldots, n-1 \}$.
Index the set of possible pairs by $(i,j) \in [n]^2$ with $i < j$.
Label rows and columns of the large square using numbers from $[n]$.
When $n$ is odd, place the pair $(i,j)$ at row $k$ of the large square where $i + j \equiv k \pmod n$.
If two pairs $(i_1,j_1)$, $(i_2,j_2)$ on same row intersect, then
one of the following happens
$$i_1 = i_2 \lor i_1 = j_2\lor j_1 = i_2 \lor j_1 = j_2$$
Since $i_1 + j_1 \equiv i_2 + j_2 \pmod n$, we find
$$(i_1,j_1) = (i_2,j_2) \pmod n \lor (i_1,j_1) = (j_2,i_2) \pmod n$$
Since $i_1,i_2,j_1,j_2 \in [n]$ and $i_1 < j_1$, $i_2 < j_2$, we can rule out the second case. From this, we can deduce distinct pairs on some row are disjoint. This generate
a desired packing of the $\binom{n}{2}$ pairs into a $n \times n$ square.
When $n$ is even, $n - 1$ is odd.
Place those pair $(i,j) \in [n-1]^2$ into row $k$ where $i + j \equiv k \pmod {n-1}$.
Notice
- For each row $i \in [n-1]$, the slot at column $2i \pmod {n - 1}$ and $n-1$ is unused.
- For any column $j \in [n-1]$, one and only slot at row $i \in [n-1]$ is unused.
For those pair $(i,j) \in [n]^2 \setminus [n-1]^2$ with $i < j$, we have $j = n$.
We can place the pair on row $k$ where $2k = i \pmod {n-1}$. This will fill all the unused slots in the first $n-1$ rows and generate a desired packing of the $\binom{n}{2}$ pairs into a $(n-1) \times n$ rectangle.
