-- в базе хранятся версии кучи значений, необходимо по одному id соединить все значения j последовательно
with repository_jsonb as (select 1::bigint as version, true as s, '{"id": "21843927", "id_pac": null, "org_id": "139"}'::jsonb as j
union all select 2, true, '{"id": "21843927", "org_id": "140"}'
union all select 3, true, '{"id": "21843927", "org_id": "141"}'
union all select 1, true, '{"id": "21843920", "id_pac": null, "org_id": "139"}'
union all select 4, true, '{"id": "21843927", "id_pac": "123"}'
union all select 5, true, '{"id": "21843927", "org2_id": "100"}'
),
repository_jsonb

r as (
with recursive t as (
	select j->>'id' as id, j, s, version from repository_jsonb where s and version=1
	union 
	select t.j->>'id', t.j||j.j, j.s, j.version from t inner join repository_jsonb j on id = j.j->>'id' and t.version+1=j.version where j.s
	)
select t.* from t inner join (select id, max(version) as version from t group by id) t2 using(id, version)
	)
-- рабочий вариант:
select * from r

--неудавшая попытка так как учитывается состояние только до изменения
--select *, (lead(j,-1,'{}'::jsonb) over(order by j->>'id' , version))||j, (lead(j,-1,'{}'::jsonb) over(order by j->>'id' , version)), j from 

with repository_jsonb as (select 1::bigint as version, true as s, '{"id": "21843927", "id_pac": null, "org_id": "139"}'::jsonb as j
	union all select 2, true, '{"id": "21843927", "org_id": "140"}'
	union all select 3, true, '{"id": "21843927", "org_id": "141"}'
	union all select 1, true, '{"id": "21843920", "id_pac": null, "org_id": "139"}'
	union all select 4, true, '{"id": "21843927", "id_pac": "123"}'
	union all select 5, true, '{"id": "21843927", "org2_id": "100"}'
	)
	select id, version, key, value from (
	SELECT (j->>'id')::bigint as id, version, max(version) over(partition by j->>'id', t.key) as max_version , t.key, t.value FROM repository_jsonb, LATERAL jsonb_each_text(J) T 
	)t where version = max_version