One rainy afternoon you write the number $1$ once, the number $2$ twice, the number $3$ three times, and so forth until you have written the number $99$ ninety-nine times. What is the $2005$ th digit that you write?
1. We need to determine the 2005th digit in the sequence where the number \( n \) is written \( n \) times for \( n = 1, 2, 3, \ldots, 99 \).

2. First, we calculate the total number of digits written for each number \( n \):
   - For single-digit numbers (1 to 9), each number \( n \) is written \( n \) times, contributing \( n \) digits each.
   - For two-digit numbers (10 to 99), each number \( n \) is written \( n \) times, contributing \( 2n \) digits each.

3. Calculate the total number of digits contributed by the single-digit numbers:
   \[
   \sum_{n=1}^{9} n = 1 + 2 + 3 + \ldots + 9 = \frac{9 \cdot 10}{2} = 45 \text{ digits}
   \]

4. Next, we need to find the number of digits contributed by the two-digit numbers. We need to find the smallest \( n \) such that the cumulative number of digits reaches or exceeds 2005.

5. Let \( S \) be the cumulative number of digits written up to number \( n \):
   \[
   S = 45 + \sum_{n=10}^{k} 2n
   \]

6. We need to find \( k \) such that:
   \[
   45 + \sum_{n=10}^{k} 2n \leq 2005
   \]

7. Simplify the sum of the two-digit numbers:
   \[
   \sum_{n=10}^{k} 2n = 2 \sum_{n=10}^{k} n = 2 \left( \frac{k(k+1)}{2} - \frac{9 \cdot 10}{2} \right) = k(k+1) - 90
   \]

8. Therefore, we need:
   \[
   45 + k(k+1) - 90 \leq 2005
   \]
   \[
   k(k+1) - 45 \leq 2005
   \]
   \[
   k(k+1) \leq 2050
   \]

9. Solve the quadratic inequality:
   \[
   k^2 + k - 2050 \leq 0
   \]

10. Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = 1 \), and \( c = -2050 \):
    \[
    k = \frac{-1 \pm \sqrt{1 + 8200}}{2} = \frac{-1 \pm \sqrt{8201}}{2}
    \]
    \[
    k \approx \frac{-1 + 90.55}{2} \approx 44.77
    \]

11. Since \( k \) must be an integer, we take \( k = 44 \).

12. Verify the number of digits:
    \[
    S = 45 + \sum_{n=10}^{44} 2n = 45 + 2 \left( \frac{44 \cdot 45}{2} - \frac{9 \cdot 10}{2} \right) = 45 + 2(990 - 45) = 45 + 1890 = 1935
    \]

13. We need 70 more digits to reach 2005. The next number is 45, which is written 45 times, contributing 90 digits.

14. The 70th digit in the sequence of 45s is the 25th digit of the 45th number, which is 5.

The final answer is \( \boxed{5} \).