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In this math.stackexchange.com question I seek a tighter bound than the one I presented in there in the question. Rob Pratt puts forth a conjecture in his answer motivated by the dual problem of the relaxation of the integer linear program. Can anyone either present a righter bound or provide a proof to Rob Pratt's conjecture?

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    $\begingroup$ Readers should note that Hans uses a nonstandard definition of Latin square, in which each number appears once in each diagonal as well as in each row and each column. $\endgroup$
    – Gerry Myerson
    Commented Aug 31, 2019 at 1:53

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For each $1\le i\le n-1$ the sum of numbers of $i$-th column of the lower triangle $\Delta$ below the main diagonal is at least $1+\dots+i=i(i+1)/2$. Thus the sum of all numbers in $\Delta$ is at least $\sum_{i=1}^{n-1} i(i+1)/2=(n-1)n(n+1)/6$. It easy to see that this bound is not tight for each $n\ge 3$.

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