Let $T = \{9^k : k \ \text{is an integer}, 0 \le k \le 4000\}$.  Given that $9^{4000}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of $T$ have 9 as their leftmost digit?
1. **Understanding the Problem:**
   We need to determine how many elements of the set \( T = \{9^k : k \ \text{is an integer}, 0 \le k \le 4000\} \) have 9 as their leftmost digit. We are given that \( 9^{4000} \) has 3817 digits and its first digit is 9.

2. **Number of Digits in \( 9^k \):**
   The number of digits \( d \) of a number \( n \) can be found using the formula:
   \[
   d = \lfloor \log_{10} n \rfloor + 1
   \]
   For \( 9^k \), we have:
   \[
   d = \lfloor \log_{10} (9^k) \rfloor + 1 = \lfloor k \log_{10} 9 \rfloor + 1
   \]
   Given \( 9^{4000} \) has 3817 digits, we can write:
   \[
   3817 = \lfloor 4000 \log_{10} 9 \rfloor + 1
   \]
   Solving for \( \log_{10} 9 \):
   \[
   3816 = \lfloor 4000 \log_{10} 9 \rfloor
   \]
   \[
   \frac{3816}{4000} \leq \log_{10} 9 < \frac{3817}{4000}
   \]
   \[
   0.954 \leq \log_{10} 9 < 0.95425
   \]

3. **Condition for Leftmost Digit to be 9:**
   For \( 9^k \) to have 9 as its leftmost digit, the fractional part of \( k \log_{10} 9 \) must satisfy:
   \[
   0.954 \leq \{ k \log_{10} 9 \} < 1
   \]
   where \( \{ x \} \) denotes the fractional part of \( x \).

4. **Counting the Valid \( k \):**
   The fractional part \( \{ k \log_{10} 9 \} \) is uniformly distributed over the interval \([0, 1)\). The length of the interval \([0.954, 1)\) is:
   \[
   1 - 0.954 = 0.046
   \]
   Therefore, the proportion of \( k \) values for which \( 9^k \) has 9 as the leftmost digit is 0.046.

5. **Calculating the Number of Valid \( k \):**
   The total number of \( k \) values is 4001 (from 0 to 4000 inclusive). Thus, the number of \( k \) values for which \( 9^k \) has 9 as the leftmost digit is:
   \[
   0.046 \times 4001 \approx 184
   \]

Conclusion:
\[
\boxed{184}
\]