程序已修改，为测验程序，增加了数据复杂度。
经测试，下面程序可以使用。

data test;
input subjn dt Rx $ x;
cards;
1 1 A 11
1 2 B 12
1 3 C 13
2 3 A 21
2 7 C 22
2 7 B 23
2 12 D 24
2 12 B 25
3 1 B 31
3 1 C 32
3 5 A 33
3 5 C 58
3 8 B 34
3 11 B 35
3 11 D 36
3 11 E 37
4 1 A 41
4 2 B 42
4 2 D 43
4 3 C 44
;run;

%macro test();
proc sql;
  drop table t1;
drop table t7;
quit;
proc sql;
  create table t1 as
select distinct subjn
  from test;
quit;
data _null_;
  set t1 nobs=n;
call symput('n', n);
stop;
run;
%do ii = 1 %to &n.;
data _null_;
  set t1(firstobs=&ii. obs=&ii.);
  call symput('subjn', subjn);
  stop;
run;
data t2;
  set test;
  where subjn = &subjn.;
run;
proc sort data = t2;
  by dt rx;
run;
proc sql;
  create table t2_only as
  select *
   from t2
   where dt in (select distinct dt from ( select dt, count(*) as cnt from t2 group by dt) where cnt = 1 );
quit;
proc sql;
  create table t2_else as
  select *
   from t2
   where dt not in (select dt from t2_only)
   order by dt;
quit;
data t2_else;
  set t2_else end=eof;
  by dt;
  if _n_ = 1 then fake_id1 = 0;
  if first.dt then fake_id1 + 1;
  if eof then call symput('sql_join_times', fake_id1-1);
run;
proc sql noprint;
  select count(*) into:flg1 from t2_else;
quit;
%if &flg1. ne 0 %then %do;
  proc sql noprint;
   select count(distinct fake_id1) into:flg2 from t2_else;
  quit;
  %if &flg2. = 1 %then %do;
   data t3_1(keep=subjn dt_1 rx_1);
    set t2_else(rename=(dt=dt_1 rx=rx_1));
    output t3_1;
   run;
  %end;
  %do jj = 1 %to &sql_join_times.;
   %let name_j1 = %eval(&jj.+0);
   %let name_j2 = %eval(&jj.+1);
   %if &jj. = 1 %then %do;
    data t3_1 t3_2(drop = fake_id1);
     set t2_else;
     where fake_id1 in (1,2);
     if fake_id1 = 1 then output t3_1;
     else output t3_2;
    run;
    proc sql;
     create table t4 as
     select a.subjn, a.dt as dt_1, a.rx as rx_1, b.dt as dt_2, b.rx as rx_2
      from t3_1 as a
      full join t3_2 as b
      on a.subjn = b.subjn and a.dt < b.dt;
    quit;
   %end;%else %do;
    proc sql;
     create table t4 as
     select a.*, b.dt as dt_&name_j2., b.rx as rx_&name_j2.
      from t3_1 as a
      full join t3_2 as b
      on a.subjn = b.subjn and a.dt_&name_j1. < b.dt;
    quit;
   %end;
   data t3_1;
    set t4;
   run;
   %if &jj. ne &sql_join_times. %then %do;
    data t3_2;
     set t2_else;
     where fake_id1 = &jj. + 2;
    run;
   %end;
  %end;
  data t5_else(keep = subjn dt rx grp);
   set t3_1 end=eof;
   grp + 1;
   %do jj = 1 %to %eval(&sql_join_times.+1);
    dt = dt_&jj.;
    rx = rx_&jj.;
    output;
   %end;
   if eof then call symput('grp_no', grp);
  run;
  data t5_only(keep = subjn dt rx grp);
   set t2_only;
   do grp = 1 to &grp_no.;
    output;
   end;
  run;
  data t6;
   set t5_only
    t5_else;
  run;
%end;%else %do;
  data t6;
   set t2_only(keep = subjn dt rx);
   grp = 1;
  run;
%end;
proc sort data = t6;
  by subjn grp dt rx;
run;
proc append base = t7 data = t6; run;
proc sql;
  drop table t2;
  drop table t2_only;
  drop table t2_else;
  drop table t3_1;
  drop table t3_2;
  drop table t4;
  drop table t5_only;
  drop table t5_else;
  drop table t6;
quit;
%end;
%mend;
%test();

不只求答案，更求各种方法，望各位高手赐教

佩服对hash和sort强大的应用! 要好好学习一下。

对不起楼主，我的思路有问题，2*3的时候ok，2*2的时候结论是错的，容我晚上在好好想想~~

mark，等明白再回来看看