tl;dr:  Two possible boxes were sold, the one with 59 or the one with 8; and if it was the 8, there are two partitions of the remaining boxes that work.

The box counts are [4, 8, 27, 31, 34, 35, 44, 45, 51, 59], the sum of which is 338 which is 2mod3.

Therefore the sold box must also contain 2mod3 so that the remainder boxes have a chance of being divided into 3 equal counts.  

So [8, 35, 44, 59] are the candidates for being the sold box.

And then the sum of each color will be, in order:  [110.0, 101.0, 98.0, 93.0]

For each candidate we have to find out if, for example

[4, 27, 31, 34, 35, 44, 45, 51, 59] can be partitioned into 3 subsets each of which sums to 110.

We can subdivide the original set according to mod3 value:

0mod3: [27, 45, 51]

1mod3: [4, 31, 34]

2mod3: [8, 35, 44, 59]

So no matter which 2mod3 candidate is chosen, the remaining 9 boxes will include 3 of each mod group.

If 59 is sold, we need the sums to be 93 which is 0mod3.  So a color group will likely be one from each mod group to be 0mod3.  Example:  51+34+8 = 93  45+4+44 = 93  27+31+35 = 93  done and done  And if the answer is unique, this must be it.

A computer search shows the following ways of partitioning the remaining 9 boxes into 3 groups such that the total balls in each group will be equal.

Case 1.  Box 59 Sold:  (4, 44, 45) (8, 34, 51) (27, 31, 35) each sums to 93

Case 2.  Box  8 Sold:  (51, 59) (31, 35, 44) (4, 27, 34, 45) each sums to 110

Case 3.  Box  8 Sold:  (51, 59) (31, 34, 45) (4, 27, 35, 44) each sums to 110

----------  code  ----------

from itertools import combinations

myList=([4, 8, 27, 31, 34, 35, 44, 45, 51, 59])

for e in myList:

    shList = myList[:]

    shList.remove(e)

    target = sum(shList)/3

    if target%1 != 0:

        continue

    target = int(target)

    print('\n',e, shList,target)

    for i in range(2,6):

        for comb in combinations(shList,i):

            if sum(comb) == target:

                print(e, target, comb)