Sketch of a heuristic algorithm

I have no exact solution for this problem. But since Raphael's comment suggests it looks like the partition problem, for which heuristic algorithms have been developed, I will try a heuristic approach. This is only a sketch of a heuristic algorithm.

Let $v$ be the number of values, and let $n$ be the number of elements in the list. We assume without loss of generality that the values are the numbers in $[1..n]$. For each value $i$ we note $n_i$ the number of its occurrences in the list.

We first note that the sum of the distances for all occurrences of a value is $n$. So the sum of all distances for all elements of the list is $vn$. Hence, the average distance for all elements of the list is $vn/n$, i.e., $v$.

Our purpose is to minimize the standard deviation on distances, which amounts to minimizing the sum of the squares of distance deviations, i.e. of differences between $v$ and each distance.

When placing the occurrences of one value, we try to equalize distances, up to integer constraints (and to the fact that slots may be occupied at some point). That is how they contribute the least to the standard deviation. If we are forced to increase a distance, it is always better to increase one that deviates the least. The the optimal distance for a value $i$ is $n/n_i$. Taking into account integer constraints, $n\mod n_i$ occurrences should use the integer value just above $n/n_i$, and the others just below.

That will guide our algorithm.

But first, we note that singleton values (occurring only once) will always have the same associated distance $n$. Hence their placement does not matter and can be ignored by the algorithm. They will just take whatever slots are left available at the end.

Then, since those distances that deviate most have to be the most exact to contribute less to the sum of squares, we try to place first the values that necessarily deviate most, i.e. the values $i$ such that $|n/n_i -v|$ is greatest.

It may be a value with very many of very few occurrences at first. I think it does not actually make a difference, since the constraints created by occupying slots are in proportion of the number of values well (?) placed.

Thr first value considered can be placed without any constraint. Then The other values must be placed so as to minimize their contribution to the standard deviation, but only in the slots left free by whatever values have been placed before.

The placement of the occurrences of a value in the remaining slots can be done with a dynamic programming algorithm, so as to merge computations that place the same number of values between two positions, keeping only those that have minimal contribution to the standard deviation (i.e. minimum value for the sum of the square of their deviations).

Occasionally, there will be several minimal solutions. In that case you try to preserve some slack by choosing the minimal solution that has the remaing slots most evenly distributed. THis can be computed, for each solution, by computing the standard deviation of the distances between the remaining free slots (with repect to their mean value, not with respect to $v$).

Then you repeat for the next remaining value $j$ such that $|n/n_j -v|$ is greatest, an so on until all non singleton values are placed.

Then you put the singleton values in the remaining slots.

I believe this should generally give reasonable solution, but I have yet no idea on how to prove it or estimate the gap with an optimal solution.