I believe this is what you're after: SQL Fiddle Example

select *
from HISTORICO
where FECHA in 
(
  select FECHA
  from HISTORICO
  group by FECHA
  having count(distinct CODIGO) = (
    select count(distinct CODIGO) 
    from HISTORICO
  )
)
-- update per comments
and CODIGO = 'AAA'
-- end of update
order by FECHA, CODIGO

Explanation

The innermost query figures out how many distinct values for CODIGO exist in the table. Since you only want records where every existing value of CODIGO has a matching value of FECHA we need to know how many CODIGO values there are in total.

The next layer up returns all values of FECHA for which every value of CODIGO exists; it does this by grouping all FECHA records then counting the number of different (distinct) values of CODIGO are in that set, and comparing that to the number we calculated earlier.

The outermost query returns all records which have this FECHA value; since those are the ones we care about.

Note: there may be more efficient solutions to this out there; I've just knocked up a first pass at what should meet your functional requirement. If you have an index of FECHA (idealy your primary key & clustered index contains FECHA followed by CODIGO) this should perform pretty well.

Your Requirements

Just in case there's any confusion, here's how I understood your requirement - please say if this is incorrect, as if it is this solution won't be appropriate:

You want to:

• return all records
• where for a given value of FECHA
• there exists a record with that FECHA value for every possible value of CODIGO

Note: "every possible value of CODIGO" is based on the values that exist across the HISTORICO table itself; i.e. if we were to delete the records where CODIGO='CCC' then all records would be returned, since both AAA and BBB have values with FECHA 2022 and 2023, and there would be no other values of CODIGO (based on the sample data in your screenshots.

Just a guess, but it seems you can use the WITH TIES option in concert with the window function row_number()

Select top 1 with ties *
 From YourTable
 Order by row_number() over (partition by codigo order by Fecha)

I would guess you want to look for sets that include all FECHA values.

You can do:

select * from t where codigo in (
  select codigo from t group by codigo having count(distinct fecha) = 3
)

Change 3 by the number of FECHA you want to check.