To solve the problem, we need to find all positive integers $n$ such that $f(n) = 1996$, where $f(n)$ is the maximal natural number such that $2^{f(n)}$ divides the sum:
$$\sum_{i=0}^{\left\lfloor \frac{n - 1}{2} \right\rfloor} \binom{n}{2i+1} 3^i$$

First, we express the sum in a more manageable form. We start by using the binomial theorem and properties of binomial coefficients.

1. Express the sum using roots of unity:
$$x_n = \sum_{i=0}^{\left\lfloor \frac{n-1}{2} \right\rfloor} \binom{n}{2i+1} 3^i$$
This can be rewritten using the binomial theorem for complex numbers:
$$x_n = \frac{1}{2\sqrt{3}} \left[ (1+\sqrt{3})^n - (1-\sqrt{3})^n \right]$$

2. **Simplify the expression for even and odd $n$:**
- For $n = 2k$:
$$x_{2k} = \frac{1}{2\sqrt{3}} \left[ (1+\sqrt{3})^{2k} - (1-\sqrt{3})^{2k} \right]$$
Using the fact that $(1+\sqrt{3})^2 = 2(2+\sqrt{3})$ and $(1-\sqrt{3})^2 = 2(2-\sqrt{3})$, we get:
$$x_{2k} = \frac{2^k}{2\sqrt{3}} \left[ (2+\sqrt{3})^k - (2-\sqrt{3})^k \right]$$
Therefore,
$$x_{2k} = 2^k \sum_{i=0}^{\left\lfloor \frac{k-1}{2} \right\rfloor} \binom{k}{2i+1} 2^{k-2i-1} 3^i$$
Hence, $f(4k+2) = 2k+1$ and $f(4k) = 2k+1 + \text{ord}_2(k)$.

- For $n = 2k+1$:
$$x_{2k+1} = \frac{1}{2\sqrt{3}} \left[ (1+\sqrt{3})^{2k+1} - (1-\sqrt{3})^{2k+1} \right]$$
Using the same properties, we get:
$$x_{2k+1} = \frac{2^k}{2\sqrt{3}} \left[ (2+\sqrt{3})^k (1+\sqrt{3}) - (2-\sqrt{3})^k (1-\sqrt{3}) \right]$$
Simplifying further:
$$x_{2k+1} = 2^k \left[ \sum_{i=0}^{\left\lfloor \frac{k-1}{2} \right\rfloor} \binom{k}{2i+1} 2^{k-2i-1} 3^i + \sum_{i=0}^{\left\lfloor \frac{k}{2} \right\rfloor} \binom{k}{2i} 2^{k-2i} 3^i \right]$$
Therefore, $f(n) = \left\lfloor \frac{n}{2} \right\rfloor$ if $4 \nmid n$, and $f(n) = \left\lfloor \frac{n}{2} \right\rfloor + \text{ord}_2\left(\left\lfloor \frac{n}{2} \right\rfloor\right)$ if $4 \mid n$.

3. **Determine $n$ such that $f(n) = 1996$:**
- If $n = 2k$ and $4 \mid n$:
$$f(n) = 2k + \text{ord}_2(k) = 1996$$
Let $k = 2^m \cdot t$ where $t$ is odd. Then:
$$2k + m = 1996 \implies 2(2^m \cdot t) + m = 1996$$
Solving for $k$:
$$k = 2^{1996 - m} \cdot t$$
Therefore, $n = 4k = 4 \cdot 2^{1996 - m} \cdot t$.

- If $n = 2k+1$:
$$f(n) = k = 1996 \implies n = 2 \cdot 1996 + 1 = 3993$$

Thus, the values of $n$ such that $f(n) = 1996$ are $n = 3993$ and $n = 3984$.

The final answer is $\boxed{3993 \text{ and } 3984}$.