Затем присоедините таблицу обратно к salaries
,
SELECT *
FROM salaries
JOIN (
SELECT year, MAX(pay)
FROM salaries
WHERE year IN ( 2015 , 2016 )
GROUP BY year
) AS ms
USING (year, pay);
Или вы можете попробовать использовать dense_rank()
SELECT *
FROM (
SELECT year, pay, dense_rank() OVER (PARTITION BY year ORDER BY pay) AS dr
FROM salaries
WHERE salaries.year IN ( 2015 , 2016 )
) AS ms
WHERE dr = 1;