Let $G$ be a work schedule of the fools that enables the maximum number of consecutive joyful weeks. If 2 and 4 have a common working week then every working week of 4 is also a working week of 2. Consecutively, in the schedule obtained from $G$ by shifting the working weeks of 4 by one week, all weeks that are joyful in $G$ are still joyful. Analogously, if 8 shares a working week with 2 or 4 then the working weeks of 8 can be shifted so that they would not coincide with that of 2 or 4 while no joyful week would be lost. Thus, without loss of generality, assume that the working weeks of 2, 4 and 8 do not coincide in $G$. Then precisely every 8th week has the property that noone of 2, 4 and 8 is working.
As 3 and 8 are coprime, 3 covers every 3rd of the weeks during which noone of 2, 4 and 8 is working. Analogously to what was done above, assume without loss of generality that the working weeks of 6 do not coincide with that of 2 or 3. Then also 6 covers every 3rd of the weeks during which noone of 2, 4 and 8 is working and precisely every 24th week has the property that noone of 2, 3, 4, 6 and 8 is working. Call these weeks suspicious.
Analogously to that was done above, assume without loss of generality that 9 covers some suspicious week. As $\text{lcm}(24, 9) = 24 \cdot 3$, he covers every 3rd suspicious week. Also assume that 5 and 10 do not have common working weeks. As $\text{lcm}(24, 5) = \text{lcm}(24, 10) = 24 \cdot 5$, each of them covers every 5th suspicious week. Also, 7 and 11 cover every 7th and every 11th suspicious week, respectively. Hence we are facing a subproblem that considers only the suspicious weeks and the schedule is made for one 3, two 5s, one 7 and one 11.
The relative position of the working weeks of one 3 and one 5 is not important (by Chinese remainder theorem). By symmetry, there are only two in principle different ways to insert the second 5: he is shifted with respect to the other 5 either by one or by two suspicious weeks. Correspondingly, we get the following schedules (where $\bullet$ denotes suspicious weeks during which either 3 or one of the 5s is working and $\circ$ denotes the remaining suspicious weeks):
$$(1) \bullet\bullet\bullet\circ\circ\bullet\circ\circ\bullet\circ\circ\bullet\circ\circ\bullet\circ\circ\bullet\circ\circ\bullet\circ\circ\bullet\circ\circ\bullet\circ\circ\dots$$
$$(2) \bullet\circ\circ\circ\circ\circ\bullet\circ\circ\circ\bullet\circ\circ\circ\bullet\circ\circ\circ\bullet\circ\circ\circ\dots$$
As 7 and 15 are coprime, each of the schedules (1) and (2) gives rise to only one schedule with 7 (by Chinese remainder theorem again). These are

respectively.

Adding 11 to schedule (1') leads to maximum 8 consecutive covered suspicious weeks, but adding it to schedule (2') gives maximum 11 of them. Thus, the maximum distance of two uncovered suspicious weeks is $24 \cdot 12 = 288$ weeks and the corresponding number of consecutive joyful weeks is 287. Then the king can avoid feeling bored during $287 \cdot 7 = 2009$ days.