Luckily seven is a prime, so all the professors will need to skip every seventh lecture. Let the coming Sunday be number zero, let $i$ be the Sunday $i$ weeks into the future, and let $N$ denote the number of a Sunday, on which they all need to skip a class. Prof. Monday will need to skip this Sunday and every second Sunday thereafter, so $N\equiv 0 \pmod 2$. Prof. Tuesday would first like to preach on Sunday $\#1$ exactly $12=4\cdot3$ days after his course commences, so $N\equiv 1\pmod 3$. Prof. Wednesday will miss an opportunity on the coming Sunday as well, so $N\equiv 0\pmod 4$. The eager beaver prof. Thursday will be irritated on every single Sunday, and can thus be left out of the reckoning. The lazy professor Friday will first be annoyed (or relieved?) on Sunday $\#4$ (four full weeks plus two days from Fri to Sun = $30=5\cdot 6$ days), so $N=4\pmod 6$. Professor Saturday's first miss is $15=3\cdot5$ days after his course has started on Sunday $\#2$, so $N\equiv 2\pmod 5.$
BUT I'm not going to solve this for you. Check the above, and then use the CRT to find the possible values of $N$. For extra credit you should observe that it was not a priori clear that the six professors would all miss a lecture on the same Sunday. A number of miraculous fits occurred, so I give some due credit to whoever composed this problem.
Use modular arithmetic. Let's say the first day of the year is day $x$, which takes a value from $0$ to $6$, with and $0 \equiv \text{ Monday}$, so that $x$ is the number of days after the start of the first week that the transition to the new year occurs.
Now, to find what day of the week the $y$th day of the year will be, note that $y-1$ days have passed between the $1$st an $y$th day, so add on an extra $y-1$ days to whatever the first day was ($x$).
$$(y-1)+x \mod 7$$
where $... \mod 7$ is the remainder left over when you take away $7$ the maximum number of times from $...$ (that still leaves a positive integer).
In the first case, $x= \text{ Monday }=0$ and $y=260$, so
$$(260-1)+0 \mod 7 \equiv 259-7-7-7-... \mod 7 $$$$259-7(w) \mod 7$$$$\equiv 0 \mod 7$$
Which is Monday
If Wednesday is the first day, then let $x=2$, so the $260$th day is
$$(260-1)+2 \mod 7 \equiv 2 \mod 7$$
Edit: the answer with less modular arithmetic.
If the $1$st day is Monday, then the $8$th, $15$th, $22$nd... days will also be Monday. These numbers have the general form $7w+1$.
If the $1$st day is Monday, then the $2$nd, $9$th, $16$th, $23$rd... days will be Tuesday. These numbers have the general form $7w+2$.
The $3$rd, $10$th, $17$th, $24$th... days will be Wednesday. These numbers have the general form $7w+3$.
In general, why do Mondays have the form $7w+1$? Day $1$ is a Monday, and any day a multiple of $7$ up from that will also be a Monday (or, will $\equiv 1 \mod 7$). So you can add $7$ to $1$ as many times as you want without affecting the resulting number's Mondayness.
If the first day is a Wednesday, all days that year will be shifted down $2$ from the Monday situation. Thus Monday has the form $(7w+1)-2=7w-1=7w+6$, Tuesday $7w$, W. $7w+1$, Th. $7w+2$, F. $7w+3$, Sa. $7w+4$, and Sunday $7w+5$.
Best Answer
The 200th day of Year $N+1$ is either 365-100=265 or 366-100=266 days after the 300th day of Year $N$. This is a whole number of weeks. So was a Leap Year involved?
How many days earlier was the 100th day of Year $N-1$?