SQL:conta todos os registros com ocorrência consecutiva de mesmo valor para cada conjunto de dispositivos e retorna a contagem mais alta

Esta é uma forma de lacunas e ilhas. Você pode usar uma diferença de números de linha para obter as ilhas:

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);

Então, para obter o máximo, use outra camada de funções da janela:

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;