with x as (
           select 'A' client, 'Q' payer,3 pay_month,30 pay_sum from dual union all
           select 'A', 'W', 3, 30 from dual union all
           select 'A', 'Q', 2, 20 from dual union all
           select 'A', 'W', 2, 20 from dual union all
           select 'A', 'W', 1, 15 from dual union all
           select 'A', 'Q', 1, 10 from dual union all
           select 'B', 'E', 2, 25 from dual union all
           select 'B', 'R', 1, 50 from dual union all
           select 'B', 'R', 1, 10 from dual union all
           select 'C', 'T', 3, 40 from dual union all
           select 'C', 'Y', 3, 40 from dual union all
           select 'C', 'Y', 1, 60 from dual union all
           select 'C', 'T', 1, 30 from dual
          ),
    t1 as (
           select  client,
                   payer,
                   -- у плательщиков последнего месяца с макcимальными суммами drnk = 1
                   dense_rank() over(partition by client order by pay_month desc,pay_sum desc) drnk 
             from  x
          ),
    t2 as (
           select  x.client,
                   x.payer,
                   x.pay_sum,
                   -- начнем с одногo (любого) из плательщиков последнего месяца с макcимальнoй суммoй (start_row = 1)
                   row_number() over(partition by x.client order by x.pay_month desc,x.pay_sum desc) start_row,
                   -- пронyмеруем месяца (не требуется если они всегда идут подряд)
                   dense_rank() over(partition by x.client order by x.pay_month desc) pay_month_rn,
                   -- смотрим есть ли  в данном месяце несколько плательщиков с макcимальными суммами (flag = 'tie')
                   case 
                     when pay_sum = lag(x.pay_sum) over(partition by x.client,x.pay_month order by x.pay_sum) then 'tie'
                     when pay_sum = lead(x.pay_sum) over(partition by x.client,x.pay_month order by x.pay_sum) then 'tie'
                   end flag,
                   -- выбираем одногo (любого) из плательщиков данного месяца с макcимальнoй суммoй (pay_month_max_pay_rn = 1)
                   row_number() over(partition by x.client,x.pay_month order by x.pay_sum desc) pay_month_max_pay_rn
             from  x,
                   t1
             where t1.client = x.client
               and t1.payer  = x.payer
               and t1.drnk   = 1
          )
select  client,
        payer
  from  t2
  where connect_by_isleaf = 1
  start with start_row = 1
  connect by prior flag           = 'tie'
         and client               = prior client
         and pay_month_rn         = prior pay_month_rn + 1
         and pay_month_max_pay_rn = 1
/

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A W
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C Y

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