Answer: A

Small Hint:

Two dates fall on the same weekday exactly when the number of days between them is a multiple of $7$

Big Hint:

Count the days from day $300$ of year $N$ to day $200$ of year $N+1$ to decide whether $N$ is a leap year

Solution:

From day $300$ of year $N$ to day $200$ of year $N + 1$ is $(L - 300) + 200$ days, where $L$ is the length of year $N.$ If $N$ were not a leap year, this is $265 \equiv 6 \pmod 7,$ giving a Monday, not a Tuesday. So year $N$ is a leap year, and the count is $266 = 7 \cdot 38,$ consistent with Tuesday.

Then years $N - 1$ and $N + 1$ are not leap years.

The $100$th day of year $N - 1$ precedes the Tuesday (day $300$ of year $N$) by $(365 - 100) + 300 = 565$ days. Since $565 = 7 \cdot 80 + 5,$ that day is $5$ days earlier in the week than Tuesday, which is a Thursday.

Thus, the correct answer is A.