To determine if we can order 365 cards from smallest to largest in 2000 steps, we need to analyze the given sum and the function $f(x) = \lceil \log_3 x \rceil$.

1. **Understanding the function $f(x) = \lceil \log_3 x \rceil$**:
- For $x = 3$, $\log_3 3 = 1$, so $f(3) = 1$.
- For $4 \leq x \leq 9$, $1 < \log_3 x \leq 2$, so $f(x) = 2$.
- For $10 \leq x \leq 27$, $2 < \log_3 x \leq 3$, so $f(x) = 3$.
- For $28 \leq x \leq 81$, $3 < \log_3 x \leq 4$, so $f(x) = 4$.
- For $82 \leq x \leq 243$, $4 < \log_3 x \leq 5$, so $f(x) = 5$.
- For $244 \leq x \leq 365$, $5 < \log_3 x \leq 6$, so $f(x) = 6$.

2. **Summing the values of $f(x)$**:
- We need to sum $\lceil \log_3 (i+1) \rceil$ for $i$ from 2 to 365.
- The sum can be broken down as follows:
$$\sum_{i=2}^{365} \lceil \log_3 (i+1) \rceil = \sum_{i=2}^{3} 1 + \sum_{i=4}^{9} 2 + \sum_{i=10}^{27} 3 + \sum_{i=28}^{81} 4 + \sum_{i=82}^{243} 5 + \sum_{i=244}^{365} 6$$

3. Calculating each segment:
- For $\sum_{i=2}^{3} 1$:
$$1 \cdot (3 - 2 + 1) = 1 \cdot 2 = 2$$
- For $\sum_{i=4}^{9} 2$:
$$2 \cdot (9 - 4 + 1) = 2 \cdot 6 = 12$$
- For $\sum_{i=10}^{27} 3$:
$$3 \cdot (27 - 10 + 1) = 3 \cdot 18 = 54$$
- For $\sum_{i=28}^{81} 4$:
$$4 \cdot (81 - 28 + 1) = 4 \cdot 54 = 216$$
- For $\sum_{i=82}^{243} 5$:
$$5 \cdot (243 - 82 + 1) = 5 \cdot 162 = 810$$
- For $\sum_{i=244}^{365} 6$:
$$6 \cdot (365 - 244 + 1) = 6 \cdot 122 = 732$$

4. Summing all segments:
$$2 + 12 + 54 + 216 + 810 + 732 = 1826$$

5. Comparing with the number of steps:
- The total number of steps required is 1826.
- Given that we have 2000 steps, we can indeed order the 365 cards from smallest to largest.

The final answer is True