From https://news.ycombinator.com/item?id=5779406

gistfile1.sql

-- Example Data

DROP TABLE Employees;
DROP TABLE Departments;

CREATE TABLE Departments(DepartmentID INTEGER PRIMARY KEY, Name VARCHAR);
CREATE TABLE Employees(EmployeeID INTEGER PRIMARY KEY, DepartmentID INTEGER, BossID INTEGER, Name VARCHAR, Salary INTEGER);

ALTER TABLE Employees ADD FOREIGN KEY (BossID) REFERENCES Employees(EmployeeID);
ALTER TABLE Employees ADD FOREIGN KEY (DepartmentID) REFERENCES Departments(DepartmentID);

INSERT INTO Departments(DepartmentID, Name)
                 VALUES(1, 'Exec'),
                       (2, 'Legal'),
                       (3, 'IT'),
                       (4, 'Admin'),
                       (5, 'Nobody');

INSERT INTO Employees(EmployeeID, DepartmentID, BossID, Name,    Salary)
               VALUES(         1,            1,      1, 'Chief',    100),
                     (         2,            3,      1, 'CTO',       95),
                     (         3,            2,      1, 'CFO',      100),
                     (         4,            3,      2, 'IT 1',      90),
                     (         5,            3,      2, 'IT 2',      90),
                     (         6,            4,      1, 'Adm 1',     20),
                     (         7,            4,      1, 'Adm 2',    110),
                     (         8,            3,      2, 'IT 3',      50),
                     (         9,            3,      1, 'IT 4',      60),
                     (        10,            2,      3, 'Legal 1',  110),
                     (        11,            3,      3, 'IT 5',      80),
                     (        12,            3,      1, 'IT 6',     200);


-- Answers


--    List employees (names) who have a bigger salary than their boss
SELECT a.Name FROM Employees a JOIN Employees b
  ON a.BossID = b.EmployeeId
  WHERE a.Salary > b.Salary;

--    List employees who have the biggest salary in their departments
SELECT a.Name, a.DepartmentID
  FROM Employees a JOIN
    (SELECT a.DepartmentID, MAX(Salary)
       FROM Employees a JOIN Departments b ON a.DepartmentID = b.DepartmentID
   GROUP BY a.DepartmentID) b
  ON a.Salary = b.max AND a.DepartmentID = b.DepartmentID;

--    List departments that have less than 3 people in it
SELECT DepartmentID, COUNT(Name)
  FROM Employees
  GROUP BY DepartmentID
  HAVING COUNT(Name) < 3;

--    List all departments along with the number of people there (tricky - people often do an "inner join" leaving out empty departments)
SELECT b.Name, COUNT(a.DepartmentID)
  FROM Employees a FULL OUTER JOIN Departments b ON a.DepartmentID=b.DepartmentID
  GROUP BY b.Name;

--    List employees that don't have a boss in the same department
SELECT a.Name FROM Employees a JOIN Employees b
  ON a.BossID = b.EmployeeId
  WHERE a.DepartmentID != b.DepartmentID;

--    List all departments along with the total salary there
SELECT b.Name, SUM(a.Salary)
  FROM Employees a FULL OUTER JOIN Departments b ON a.DepartmentID = b.DepartmentID
  GROUP BY b.name;

@twistedpair — indeed, my answer is pretty ugly and complicated for what should be easier... Your solution looks good, but it doesn't run on my Postgres 9.2.4:

florent=# SELECT em.EmployeeID, em.departmentId, MAX(salary) as salary
FROM employees em
GROUP BY em.departmentId;
ERROR: column "em.employeeid" must appear in the GROUP BY clause or be used in an aggregate function
LINE 1: SELECT em.EmployeeID, em.departmentId, MAX(salary) as salary
^
florent=#

Alas, I was incorrect. MAX(tuple) does not always correlate with the max value's employeeId on MySQL.

Upon further reflection, your solution is indeed the correct one. Nice work. 👍