The minimum is $T=4n$. Write $\Delta=L-R$ for the signed difference between the pan totals.

The weight $4n$ must change pans at least once. Such a move changes $\Delta$ by $8n$. If the values before and after the move both lie in $[-T,T]$, then $8n\le2T$. Thus $T\ge4n$.

We now construct a sequence that stays within $|\Delta|\le4n$. Put $M=4n+1$ and pair weight $t$ with weight $M-t$, for $1\le t\le2n$.

First, put the two members of each pair on the same pan. At any intermediate stage, write the heavier and lighter totals as $A+s$ and $A-s$, where $A$ is half the total mass. The total mass is even, so $s$ is an integer, and the imbalance bound gives $0\le s\le2n$.

A weight $w$ can legally move from the heavier pan if $w\le2n+s$, and from the lighter pan if $w\le2n-s$. These conditions follow by requiring the resulting half-difference to have absolute value at most $2n$.

Consider a split pair $(t,M-t)$. If $t$ is on the heavier pan, move it to the lighter pan; this is legal because $t\le2n$. If $t$ is on the lighter pan and $t\le2n-s$, move it to the heavier pan. Otherwise, integrality gives $t\ge2n-s+1$, so

$M-t\le2n+s.$

In this case move $M-t$ from the heavier pan to the lighter pan. Thus every split pair can be reunited in one legal move, without splitting any other pair.

After finitely many such moves, every pair is whole. Every whole pair has mass $M>4n$, so the imbalance, which is now a multiple of $M$, must be zero. There are $n$ whole pairs on each pan.

Next, reverse all the whole pairs. Use the pair $B=(2n,2n+1)$ to exchange pairs between the pans. If $P=(t,M-t)$ is on the left and $B$ is on the right, move the following weights, in this order:

$2n,\quad M-t,\quad 2n+1,\quad t.$

The successive signed imbalances are

$4n,\quad 2t-4n-2,\quad 2t,\quad 0.$

Here $1\le t\le2n-1$, so every imbalance lies in $[-4n,4n]$. This exchanges $B$ and $P$ and leaves every other pair unchanged. Reflecting the moves handles the opposite orientation.

To exchange two opposite pairs $P,Q$ neither of which is $B$, suppose $B$ shares $P$'s pan. Exchange $B$ with $Q$, then $B$ with $P$. The result exchanges $P$ and $Q$ and restores $B$ to its original pan. The reflected construction covers the other case. Therefore any two opposite whole pairs can be exchanged. Match the $n$ pairs on one pan with the $n$ pairs on the other and exchange each matched pair of pairs. Every weight has now switched pans relative to this whole-pair arrangement.

Finally, let $S$ be the original arrangement and $C$ the whole-pair arrangement reached in the first stage. For any arrangement $X$, let $X^*$ denote its reflection, with the pans exchanged. The first stage gives a legal path $S\to C$. Reflecting and reversing that path gives a legal path $C^*\to S^*$. The second stage gives $C\to C^*$. Combining them gives

$S\to C\to C^*\to S^*,$

which reverses the original arrangement while maintaining the bound throughout.

Source: IMO Shortlist 2024, C6, proposed by Ghana. This problem was shortlisted and was not used in the IMO exam.