Let $a, b, c, d$ be distinct non-zero real numbers satisfying the following two conditions:
$ac = bd$ and  $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}= 4$.
Determine the largest possible value of the expression $\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}$.

1. Given the conditions \( ac = bd \) and \( \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} = 4 \), we start by analyzing the second condition. 

2. Notice that the equality \( \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} = 4 \) suggests a symmetry. We can use the AM-GM inequality to gain insight:
   \[
   \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4 \sqrt[4]{\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} \cdot \frac{d}{a}} = 4
   \]
   Since the equality holds, it must be that:
   \[
   \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{d}{a}
   \]

3. Let \( \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{d}{a} = k \). Then:
   \[
   k^4 = 1 \implies k = 1 \text{ or } k = -1
   \]
   Since \( a, b, c, d \) are distinct and non-zero, \( k \neq 1 \). Therefore, \( k = -1 \).

4. If \( k = -1 \), then:
   \[
   \frac{a}{b} = -1, \quad \frac{b}{c} = -1, \quad \frac{c}{d} = -1, \quad \frac{d}{a} = -1
   \]
   This implies:
   \[
   a = -b, \quad b = -c, \quad c = -d, \quad d = -a
   \]

5. Substituting \( a = -b \) into \( ac = bd \):
   \[
   (-b)c = b(-c) \implies -bc = -bc
   \]
   This is always true, so the conditions are consistent.

6. Now, we need to find the value of \( \frac{a}{c} + \frac{c}{a} + \frac{b}{d} + \frac{d}{b} \):
   \[
   \frac{a}{c} = \frac{-b}{-c} = \frac{b}{c}, \quad \frac{c}{a} = \frac{-c}{-b} = \frac{c}{b}, \quad \frac{b}{d} = \frac{b}{-a} = -\frac{b}{a}, \quad \frac{d}{b} = \frac{-a}{b} = -\frac{a}{b}
   \]

7. Since \( a = -b \) and \( c = -d \), we have:
   \[
   \frac{a}{c} = \frac{-b}{-d} = \frac{b}{d}, \quad \frac{c}{a} = \frac{-d}{-b} = \frac{d}{b}, \quad \frac{b}{d} = \frac{b}{-(-b)} = \frac{b}{b} = 1, \quad \frac{d}{b} = \frac{-(-b)}{b} = 1
   \]

8. Therefore:
   \[
   \frac{a}{c} + \frac{c}{a} + \frac{b}{d} + \frac{d}{b} = \frac{b}{d} + \frac{d}{b} + 1 + 1 = 1 + 1 + 1 + 1 = 4
   \]

The final answer is \(\boxed{4}\).