How many ordered triples of nonzero integers $(a, b, c)$ satisfy $2abc = a + b + c + 4$?
To solve the problem, we need to find all ordered triples of nonzero integers \((a, b, c)\) that satisfy the equation:
\[ 2abc = a + b + c + 4. \]

1. **Rearrange the equation:**
   \[ 2abc - a - b - c = 4. \]

2. **Solve for \(c\):**
   \[ 2abc - a - b - c = 4 \]
   \[ c(2ab - 1) = a + b + 4 \]
   \[ c = \frac{a + b + 4}{2ab - 1}. \]

3. **Analyze the denominator and numerator:**
   Notice that for \(c\) to be an integer, the numerator \(a + b + 4\) must be divisible by the denominator \(2ab - 1\). Also, \(2ab - 1\) must be a divisor of \(a + b + 4\).

4. **Consider the magnitude of \(a\) and \(b\):**
   If \(|a|, |b| > 2\), then \(2ab - 1\) will be much larger than \(a + b + 4\), making \(c\) a non-integer. Therefore, at least one of \(a\) or \(b\) must be \(\pm 1\).

5. **Check cases where \(a\) or \(b\) is \(\pm 1\):**

   - **Case 1: \(a = 1\)**
     \[ c = \frac{1 + b + 4}{2(1)b - 1} = \frac{b + 5}{2b - 1}. \]
     We need \(2b - 1\) to divide \(b + 5\):
     \[ b + 5 = k(2b - 1) \]
     For integer \(k\), solve:
     \[ b + 5 = 2kb - k \]
     \[ b - 2kb = -k - 5 \]
     \[ b(1 - 2k) = -k - 5 \]
     Check integer solutions for \(b\):
     - \(b = 1\):
       \[ c = \frac{1 + 1 + 4}{2(1)(1) - 1} = \frac{6}{1} = 6 \]
       \((a, b, c) = (1, 1, 6)\)
     - \(b = -1\):
       \[ c = \frac{1 - 1 + 4}{2(1)(-1) - 1} = \frac{4}{-3} \text{ (not an integer)} \]

   - **Case 2: \(a = -1\)**
     \[ c = \frac{-1 + b + 4}{2(-1)b - 1} = \frac{b + 3}{-2b - 1}. \]
     We need \(-2b - 1\) to divide \(b + 3\):
     \[ b + 3 = k(-2b - 1) \]
     For integer \(k\), solve:
     \[ b + 3 = -2kb - k \]
     \[ b + 2kb = -k - 3 \]
     \[ b(1 + 2k) = -k - 3 \]
     Check integer solutions for \(b\):
     - \(b = -1\):
       \[ c = \frac{-1 - 1 + 4}{2(-1)(-1) - 1} = \frac{2}{1} = 2 \]
       \((a, b, c) = (-1, -1, 2)\)
     - \(b = 1\):
       \[ c = \frac{-1 + 1 + 4}{2(-1)(1) - 1} = \frac{4}{-3} \text{ (not an integer)} \]

6. **Count the valid solutions:**
   - From Case 1: \((1, 1, 6)\) and its permutations: \((1, 1, 6)\), \((1, 6, 1)\), \((6, 1, 1)\).
   - From Case 2: \((-1, -1, 2)\) and its permutations: \((-1, -1, 2)\), \((-1, 2, -1)\), \((2, -1, -1)\).

Thus, there are \(3 + 3 = 6\) valid ordered triples.

The final answer is \(\boxed{6}\).