Answer : $n(n+1)/2$
Maximum binary pairs is possible iff the poset is a toset.
A toset is reflexive so I don't have control over self loops like $(1,1),(2,2).$ They have to be there.
My approach to this was :
If we consider $n$ elements in a set, we have $n^2-n=n(n-1)$ non diagonal elements, considering a matrix view of toset. Thus, pairs possible : $n(n-1)/2$ (which will be going in antisymmetric property, well thats not the point though here).
I'm missing out some cases because final answer is "my answer $+n$" which will become sum of first $n$ natural numbers.
Please help me find out what case am I missing, or have I got this everything wrong.
I suspect that binary pair here means ordered pair belonging to the partial order. If $P$ is a partial order on $[n]=\{1,2,\ldots,n\}$, then $P$ contains all of the pairs $\langle k,k\rangle$ on the diagonal of your matrix (by reflexivity) as well as at most half of the off-diagonal pairs (by antisymmetry), so
$$|P|\le n+\frac{n(n-1)}2=\frac{n^2-n+2n}2=\frac{n^2+n}2=\frac{n(n+1)}2\;.$$