使用MAX：）获取最大值...然后基于这两个值再次加入...

SELECT ref.score 
, subsel.max_val 
, subsel.type_id
FROM (
  SELECT res.score, max(ref.value) as max_val, ref.type_id
  FROM `refernce` ref
  JOIN `result` res
      ON ref.type_id = res.type_id
  WHERE res.score >= ref.score
  GROUP BY res.score, ref.type_id
) subsel
JOIN `refernce` ref
ON ref.type_id = subsel.type_id
AND ref.value = subsel.max_val 
WHERE 1=1
ORDER BY 3 desc;

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如果你使用的是一些更复杂的数据库（Postgre，Oracle，...）
您可以使用窗口函数来分区和排序行..然后只在随后的筛选中选择您需要的行
示例（未测试）

SELECT score_ref
, value
, type_id
FROM (
  SELECT res.score as score_res
  , ref.score as score_ref
  , ref.value
  , ref.type_id
  , row_number() over (partition by ref.type_id order by ref.value desc) as row_order
  FROM `refernce` ref
  JOIN `result` res
      ON ref.type_id = res.type_id
  WHERE res.score >= ref.score
  GROUP BY ref.type_id
)
WHERE row_number = 1

型

只使用一个子查询的解决方案，来自链接reference和result数据表，其中score和value数据行会转换成固定长度的十六进制字串表示，然后加以组合，并套用max汇总函数，再将十六进制字串转换回整数。

select
  conv(substr(binstr, 1, 3), 16, 10) as score,
  conv(substr(binstr, -5), 16, 10) as `value`,
  type_id
from (
  select
    max(concat(
          lpad(conv(ref.score, 10, 16), 3, '0'),
          lpad(conv(ref.`value`, 10, 16), 5, '0')
    )) as binstr,
    ref.type_id
  from reference as ref
  join result as res
  on ref.type_id = res.type_id and
     ref.score <= res.score
  group by ref.type_id
) as t;

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测试结果：

+-------+-------+---------+
| score | value | type_id |
+-------+-------+---------+
|     2 |    30 |       1 |
|     3 |    25 |       2 |
|     0 |     8 |       3 |
+-------+-------+---------+

型
SQL小操作。