Smallest sum (and largest difference) of two 3-digit integers

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Ṣs3¬Þ€Fs2ZḌṚ

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Background

Consider six digits a < b < c < d < e < f.

If a ≠ 0, a number pair of three-digit integers with minimal sum must clearly use a and b for the leftmost digits, c and d for the middle digits, and e and f for the rightmost digits.

That gives eight possible arrangements with identical sums (100(a + b) + 10(c + d) + (e + f)).

Since the difference should be as large as possible, all digits of the first integer should be larger than the corresponding digits of the second integer, leaving bdf₁₀, ace₁₀ as the optimal arrangement (difference 100(b - a) + 10(d - c)+ (f - e)).

Finally, if a = 0, a should still occur as early as possible (as middle digit), and a similar process reveals that the pair cdf₁₀, bae₁₀ is the correct solution.

How it works

Ṣs3¬Þ€Fs2ZḌṚ  Main link. Argument: <a, b, c, d, e, f> (in any order)

Ṣ             Sort; yield [a, b, c, d, e, f].
 s3           Split into triplets; yield [[a, b, c], [d, e, f]].
   ¬Þ€        Sort each triplet by logical NOT.
              If a ≠ 0, all digits have logical NOT 0, so this leaves the triplets
              unaltered. If a = 0, its logical NOT is 1, so the first triplet is
              sorted as [b, c, a], leaving [[b, c, a], [d, e, f]].
      F       Flatten; yield [a, b, c, d, e, f] or [b, c, a, d, e, f].
       s2     Split into pairs; yield [[a, b], [c, d], [e, f]] or
              [[b, c], [a, d], [e, f]].
         Z    Zip; yield [[a, c, e], [b, d, f]] or [[b, a, e], [c, d, f]].
          Ḍ   Undecimal; convert each triplet from base 10 to integer.
           Ṛ  Reverse the order of the generated integers.