Solution:

The maximum value is $n^{2}$, which is attained for the partition into $n$ arithmetic progressions $\{1, n+1,2 n+1\}, \ldots,\{n, 2 n, 3 n\}$, each of difference $n$.

Suppose indeed that the set has been partitioned into $N$ progressions, of respective lengths $\ell_{i}$, and differences $d_{i}$, for $1 \leqslant i \leqslant N$. Since $\ell_{i} \geqslant 3$,
$$2 \sum_{i=1}^{N} d_{i} \leqslant \sum_{i=1}^{N}\left(\ell_{i}-1\right) d_{i}=\sum_{i=1}^{N} a_{i}-\sum_{i=1}^{N} b_{i}$$
where $a_{i}$ and $b_{i}$ denote, respectively, the largest and smallest elements of progression $i$. Now
$$\begin{aligned} & \sum_{i=1}^{N} b_{i} \geqslant 1+2+\cdots+N=N(N+1) / 2 \\ & \sum_{i=1}^{N} a_{i} \leqslant(3 n-N+1)+\cdots+3 n=N(6 n-N+1) / 2 \end{aligned}$$
and thus
$$2 \sum_{i=1}^{N} d_{i} \leqslant N(3 n-N) \leqslant 2 n^{2}$$
as $N(3 n-N)$ is increasing in $N$ on the interval $[0,3 n / 2]$ and since $N \leqslant n$. This completes the proof.