Calculating how many Pomodori fit into a period of time
I think it would be simpler just to apply some ingenuity and partially unroll the loop.  While it's possible to fit in a long cycle (pomodoro, short break, pomodoro, short break, pomodoro, short break, pomodoro, long break) with time left over, then do that.  While it's possible to fit in a short cycle (pomodoro, short break) with time left over, do that.

I'd also define some constants for readability.

pomodori is not quite an appropriate name for the result, since it includes pomodori and breaks.

It would be a good habit to encapsulate your code in a function.

SECOND = SECONDS = 1
MINUTE = MINUTES = 60 * SECONDS
HOUR = HOURS = 60 * MINUTES

POMODORO = 25 * MINUTES
SHORT_BREAK = 5 * MINUTES
LONG_BREAK = 15 * MINUTES

def manage(time):
    """ Devise a Pomodoro schedule for a time span, given in seconds. """
    schedule = []
    while time &gt; 4 * POMODORO + 3 * SHORT_BREAK + LONG_BREAK:
        schedule.extend(3 * ['pomodoro', 'short-break'] + ['pomodoro', 'long-break'])
        time -= 4 * POMODORO + 3 * SHORT_BREAK + LONG_BREAK
    while time &gt; POMODORO + SHORT_BREAK:
        schedule.extend(['pomodoro', 'short-break'])
        time -= POMODORO + SHORT_BREAK
    # 0 &lt; time &lt; POMODORO + SHORT_BREAK
    schedule.append('pomodoro')
    return schedule

print(manage(3 * HOURS))