SQL：デバイスセットごとに同じ値が連続して出現するすべてのレコードをカウントし、最大のカウントを返します

これはギャップと島の形です。行番号の違いを使用して、島を取得できます：

select device_id, speed, count(*) as num_times
from (select t.*,
             row_number() over (partition by device_id order by datetime) as seqnum,
             row_number() over (partition by device_id, speed order by datetime) as seqnum_s
      from t
     ) t
group by device_id, speed, (seqnum - seqnum_s);

次に、最大値を取得するには、ウィンドウ関数の別のレイヤーを使用します。

select device_id, speed, num_times
from (select device_id, speed, count(*) as num_times,
             row_number() over (partition by device_id order by count(*) desc) as seqnum
      from (select t.*,
                   row_number() over (partition by device_id order by datetime) as seqnum,
                   row_number() over (partition by device_id, speed order by datetime) as seqnum_s
            from t
           ) t
      group by device_id, speed, (seqnum - seqnum_s)
     ) ds
where seqnum = 1;