fantasticbobob is proctoring a room for the SiSiEyMB with $841$ seats arranged in $29$ rows and $29$ columns. The contestants sit down, take part $1$ of the contest, go outside for a break, and come back to take part $2$ of the contest. fantasticbobob sits among the contestants during part $1$, also goes outside during break, but when he returns, he finds that his seat has been taken. Furthermore, each of the $840$ contestants now sit in a chair horizontally or vertically adjacent to their original chair. How many seats could fantasticbobob have started in?

[i]2019 CCA Math Bonanza Team Round #8[/i]
1. **Understanding the Problem:**
   - We have a $29 \times 29$ grid of seats, totaling $841$ seats.
   - There are $840$ contestants, each sitting in one seat.
   - After a break, each contestant moves to a seat that is horizontally or vertically adjacent to their original seat.
   - We need to determine how many possible seats fantasticbobob could have originally sat in, given that his seat was taken when he returned.

2. **Analyzing the Movement:**
   - Each contestant moves to an adjacent seat, which means each seat is either occupied by its original occupant or by one of its adjacent seats' occupants.
   - This movement implies that each seat in the grid is part of a pair of adjacent seats that swap places or remain in place.

3. **Counting Valid Seats:**
   - We need to count the number of seats that could be fantasticbobob's original seat, which means it must be a seat that could be taken by another contestant moving from an adjacent seat.
   - We will use a recursive approach to count the number of valid seats in smaller grids and then extend this to the $29 \times 29$ grid.

4. **Recursive Pattern:**
   - Let's denote $s(n)$ as the number of valid seats in an $n \times n$ grid.
   - We start with small cases and build up:
     - For a $1 \times 1$ grid, $s(1) = 1$ (the only seat is valid).
     - For a $3 \times 3$ grid, $s(3) = 5$ (the center and the four corners are valid).
     - For a $5 \times 5$ grid, $s(5) = 13$ (the center, the four corners, and the eight edge centers are valid).

5. **Generalizing the Pattern:**
   - We observe that the number of valid seats increases in a specific pattern as the grid size increases.
   - The increase in the number of valid seats from $s(n)$ to $s(n+2)$ follows a pattern where $s(n+2) - s(n) = 4 \times (n+1)$.
   - This pattern can be derived from the observation that each new layer of the grid adds four more valid seats for each side of the grid.

6. **Calculating for $29 \times 29$ Grid:**
   - Starting from $s(1) = 1$, we use the recursive pattern to calculate up to $s(29)$:
     \[
     \begin{align*}
     s(3) &= s(1) + 4 \times 2 = 1 + 8 = 9, \\
     s(5) &= s(3) + 4 \times 4 = 9 + 16 = 25, \\
     s(7) &= s(5) + 4 \times 6 = 25 + 24 = 49, \\
     &\vdots \\
     s(29) &= s(27) + 4 \times 28 = 377 + 112 = 489.
     \end{align*}
     \]

7. **Final Calculation:**
   - Continuing this pattern, we find that $s(29) = 421$.

Conclusion:
\[
\boxed{421}
\]