Access Vb6 query

Consider joining aggregates of both tables separately by id:

Aggregate Queries (save as stored Access queries)

SELECT tb1.idF
     , tb1.[name]
     , SUM(tb1.Give) AS TG
FROM tblGive tb1
GROUP BY tb1.idF
       , tb1.[name] 

SELECT tb2.IDB
     , tb2.[name]
     , SUM(tb2.Have) AS TH
FROM tblHave tb2
GROUP BY tb2.IDB
       , tb2.name

Final Query (running Full Join Query to return all distinct names in either tables)

SELECT NZ(agg1.idF, agg2.idB) AS [id]
     , NZ(agg1.name, agg2.name) AS [name]
     , NZ(agg2.TH, 0) AS [Have]
     , NZ(agg1.TG, 0) AS [Give]
FROM tblGiveAgg agg1
LEFT JOIN tblHaveAgg agg2
   ON agg1.idF = agg2.idB

UNION 

SELECT NZ(agg1.idF, agg2.idB) AS [id]
     , NZ(agg1.name, agg2.name) AS [name]
     , NZ(agg2.TH, 0) AS [Have]
     , NZ(agg1.TG, 0) AS [Give]
FROM tblGiveAgg agg1
RIGHT JOIN tblHaveAgg agg2
   ON agg1.idF = agg2.idB;

To demonstrate with below data

CREATE TABLE tblGive (
   ID AUTOINCREMENT,
   IdF INTEGER,
   [Name] TEXT(10),
   Give INTEGER
);

INSERT INTO tblGive (IdF, [Name], [Give]) VALUES (1, 'JOHN', 37);
INSERT INTO tblGive (IdF, [Name], [Give]) VALUES (2, 'ANNA', 10);
INSERT INTO tblGive (IdF, [Name], [Give]) VALUES (3, 'BILL', -37);
INSERT INTO tblGive (IdF, [Name], [Give]) VALUES (2, 'ANNA', 116);
INSERT INTO tblGive (IdF, [Name], [Give]) VALUES (1, 'JOHN', 120);


CREATE TABLE tblHave (
   ID AUTOINCREMENT,
   IDB INTEGER,
   [Name] TEXT(10),
   Have INTEGER
);

INSERT INTO tblHave (IDB, [Name], [Have]) VALUES (1, 'JOHN', 200);
INSERT INTO tblHave (IDB, [Name], [Have]) VALUES (2, 'ANNA', 400);
INSERT INTO tblHave (IDB, [Name], [Have]) VALUES (3, 'BILL', 150);
INSERT INTO tblHave (IDB, [Name], [Have]) VALUES (1, 'JOHN', 25);
INSERT INTO tblHave (IDB, [Name], [Have]) VALUES (1, 'JOHN', 70);

Final Full Join Query returns following result: