Let $S$ be a planar region. A $\emph{domino-tiling}$ of $S$ is a partition of $S$ into $1\times2$ rectangles. (For example, a $2\times3$ rectangle has exactly $3$ domino-tilings, as shown below.)

[asy]
import graph; size(7cm); 
pen dps = linewidth(0.7); defaultpen(dps);

draw((0,0)--(3,0)--(3,2)--(0,2)--cycle, linewidth(2)); 
draw((4,0)--(4,2)--(7,2)--(7,0)--cycle, linewidth(2)); 
draw((8,0)--(8,2)--(11,2)--(11,0)--cycle, linewidth(2)); 

draw((1,0)--(1,2));
draw((2,1)--(3,1));
draw((0,1)--(2,1), linewidth(2));
draw((2,0)--(2,2), linewidth(2));

draw((4,1)--(7,1));
draw((5,0)--(5,2), linewidth(2));
draw((6,0)--(6,2), linewidth(2));

draw((8,1)--(9,1));
draw((10,0)--(10,2));
draw((9,0)--(9,2), linewidth(2));
draw((9,1)--(11,1), linewidth(2));

[/asy]
The rectangles in the partition of $S$ are called $\emph{dominoes}$.
(a) For any given positive integer $n$, find a region $S_n$ with area at most $2n$ that has exactly $n$ domino-tilings.

(b) Find a region $T$ with area less than $50000$ that has exactly $100002013$ domino-tilings.
### Part (a)
1. **Define the region \( S_n \)**:
   - We need to find a region \( S_n \) with area at most \( 2n \) that has exactly \( n \) domino-tilings.
   - Consider a \( 2 \times n \) rectangle. This rectangle has an area of \( 2n \).

2. **Count the domino-tilings**:
   - A \( 2 \times n \) rectangle can be tiled with \( 1 \times 2 \) dominoes in exactly \( F_{n+1} \) ways, where \( F_n \) is the \( n \)-th Fibonacci number.
   - The Fibonacci sequence is defined as \( F_1 = 1 \), \( F_2 = 1 \), and \( F_{n} = F_{n-1} + F_{n-2} \) for \( n \geq 3 \).

3. **Find \( n \) such that \( F_{n+1} = n \)**:
   - We need to find \( n \) such that \( F_{n+1} = n \).
   - For small values of \( n \), we can check:
     - \( F_1 = 1 \)
     - \( F_2 = 1 \)
     - \( F_3 = 2 \)
     - \( F_4 = 3 \)
     - \( F_5 = 5 \)
     - \( F_6 = 8 \)
     - \( F_7 = 13 \)
     - \( F_8 = 21 \)
     - \( F_9 = 34 \)
     - \( F_{10} = 55 \)
   - We see that there is no \( n \) such that \( F_{n+1} = n \) for small values of \( n \).

4. **Conclusion**:
   - For any given positive integer \( n \), we can use a \( 2 \times n \) rectangle which has an area of \( 2n \) and has \( F_{n+1} \) domino-tilings.
   - However, since \( F_{n+1} \) does not equal \( n \) for any \( n \), we need to find a different approach.

### Part (b)
1. **Define the region \( T \)**:
   - We need to find a region \( T \) with area less than \( 50000 \) that has exactly \( 100002013 \) domino-tilings.

2. **Use the given solution**:
   - The solution involves constructing a sequence of regions \( T_n \) using two operations:
     - Operation [1]: Attach a domino to the 2-unit side of the last domino.
     - Operation [2]: Attach a domino to the 1-unit side of the last domino.

3. **Construct the sequence**:
   - Start with \( T_3 \) which has 3 domino-tilings.
   - Use the sequence \( \overline{1112112221122111222111\ell1} \), where \( \ell \) consists of \( 24837 \) twos.

4. **Calculate the number of tilings**:
   - Using the rules described, we calculate the number of tilings for each \( T_n \):
     - \( d_1 = 1 \)
     - \( d_2 = 2 \)
     - \( d_3 = 3 \)
     - \( d_4 = d_3 + d_1 = 4 \)
     - \( d_5 = d_4 + d_3 = 7 \)
     - \( d_6 = d_5 + d_4 = 11 \)
     - \( d_7 = d_6 + d_4 = 15 \)
     - \( d_8 = d_7 + d_4 = 19 \)
     - \( d_9 = d_8 + d_4 = 23 \)
     - \( d_{10} = d_9 + d_8 = 42 \)
     - \( d_{11} = d_{10} + d_9 = 65 \)
     - \( d_{12} = d_{11} + d_9 = 88 \)
     - \( d_{13} = d_{12} + d_9 = 111 \)
     - \( d_{14} = d_{13} + d_{12} = 199 \)
     - \( d_{15} = d_{14} + d_{13} = 310 \)
     - \( d_{16} = d_{15} + d_{14} = 509 \)
     - \( d_{17} = d_{16} + d_{14} = 708 \)
     - \( d_{18} = d_{17} + d_{14} = 907 \)
     - \( d_{19} = d_{18} + d_{14} = 1106 \)
     - \( d_{20} = d_{19} + d_{18} = 2013 \)
     - \( d_{21} = d_{20} + d_{19} = 3119 \)
     - \( d_{22} = d_{21} + d_{20} = 5032 \)
     - From now on, \( d_n = d_{n-1} + d_{20} \).

5. **Final calculation**:
   - \( d_{22+24837-1} = d_{22} + 24836d_{20} = 50000000 \)
   - \( d_{22+24837} = d_{22} + 24837d_{20} = 50002013 \)
   - \( d_{24860} = 100002013 \)

6. **Area calculation**:
   - The region \( T_{24860} \) has an area of \( 2 \times 24860 = 49720 \), which is less than \( 50000 \).

Conclusion:
\[
\boxed{T_{24860}}
\]