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The longest (descending) chain starts obviously with the partition $n$ having a unique part and ends with the partition $1+1+\dots+1$ into singletons. One should move from $n$ to $1+\dots+1$ by moving dots (unities) down row by row whenever possible along staircases: \begin{eqnarray*}&&n,(n-1)+1,(n-2)+2,(n-2)+1+1,(n-3)+2+1,(n-4)+3+1,\\&&(n-4)+2+2,(n-4)+2+1+1,(n-5)+3+1+1,(n-5)+2+2+1,\\&&(n-6)+3+2+1,(n-7)+4+2+1,\dots\end{eqnarray*} etc. Rule: $\dots+\lambda_k+\lambda_{k+1}+\dots$ moves to $\dots+(\lambda_{k}-1)+ (\lambda_{k+1}+1)+\dots$ if $k$ is the largest index such that $\lambda_{k+1}-\lambda_k\geq 2$ (where $\lambda_{k+1}$ can be zero). If no such $k$ exists, diminish the last maximal part by $1$ moving the dot by the minimal possible amount (such moves are necessary in the lower half of the chain in order to "empty" the staircase). I guess one can get a formula for the length of this chain (which can probably be shown to be of maximal length by an induction argument) in terms of triangular numbers.