я есть образец таблицы ввода как

Declare  @input TABLE(Name VARCHAR(8))
INSERT INTO @input(Name)  values('Aryan') 
INSERT INTO @input(Name)  values('Aryan') 
INSERT INTO @input(Name)  values('Joseph') 
INSERT INTO @input(Name)  values('Vicky') 
INSERT INTO @input(Name)  values('Jaesmin') 
INSERT INTO @input(Name)  values('Aryan') 
INSERT INTO @input(Name)  values('Jaesmin') 
INSERT INTO @input(Name)  values('Vicky') 
INSERT INTO @input(Name)  values('Padukon') 
INSERT INTO @input(Name)  values('Aryan') 
INSERT INTO @input(Name)  values('Jaesmin') 
INSERT INTO @input(Name)  values('Vick') 
INSERT INTO @input(Name)  values('Padukon') 
INSERT INTO @input(Name)  values('Joseph') 
INSERT INTO @input(Name)  values('Marya') 
INSERT INTO @input(Name)  values('Vicky')

Также у меня есть таблица подсчета как под

declare @t table(n int)
insert into @t select 1 union all select 2 union all
select 3 union all select 4 union all select 5 union all
select 6 union all select 7 union all select 8 union all
select 9 union all select 10 union all select 11 union all
select 12 union all select 13 union all select 14 union all
select 15 union all select 16 union all select 17 union all
select 18 union all select 19 union all select 20

В Sql Server 2005, если я делаю как

 Select rn, name from (
    select ROW_NUMBER()over (order by Name) as rn , * from @input) x
    where rn % 2 <> 0

Я получаю вывод как

rn  name
1   Aryan
3   Aryan
5   Jaesmin
7   Jaesmin
9   Joseph
11  Padukon
13  Vick
15  Vicky

Бу я ограничен Sql server 2000. Как я могу получить тот же вывод?

Я пробовал с

SELECT name, (SELECT COUNT(*) FROM @input AS i2 WHERE i2.Name <= i1.Name) As rn
FROM @input AS i1

но вывод неверный

name    rn
Aryan   4
Aryan   4
Joseph  9
Vicky   16
Jaesmin 7
Aryan   4
Jaesmin 7
Vicky   16
Padukon 12
Aryan   4
Jaesmin 7
Vick    13
Padukon 12
Joseph  9
Marya   10
Vicky   16