The number 1 is a solution of the equation  

$(x + a)(x + b)(x + c)(x + d) = 16$,  

where $a, b, c, d$ are positive real numbers. Find the largest value of $abcd$.
1. Given the equation \((x + a)(x + b)(x + c)(x + d) = 16\) and knowing that \(x = 1\) is a solution, we substitute \(x = 1\) into the equation:
   \[
   (1 + a)(1 + b)(1 + c)(1 + d) = 16
   \]

2. We aim to find the maximum value of \(abcd\). To do this, we will use the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality). The AM-GM Inequality states that for any non-negative real numbers \(x_1, x_2, \ldots, x_n\):
   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}
   \]
   with equality if and only if \(x_1 = x_2 = \cdots = x_n\).

3. Applying the AM-GM Inequality to the terms \(1 + a, 1 + b, 1 + c, 1 + d\):
   \[
   \frac{(1 + a) + (1 + b) + (1 + c) + (1 + d)}{4} \geq \sqrt[4]{(1 + a)(1 + b)(1 + c)(1 + d)}
   \]

4. Since \((1 + a)(1 + b)(1 + c)(1 + d) = 16\), we have:
   \[
   \frac{(1 + a) + (1 + b) + (1 + c) + (1 + d)}{4} \geq \sqrt[4]{16}
   \]
   \[
   \frac{(1 + a) + (1 + b) + (1 + c) + (1 + d)}{4} \geq 2
   \]
   \[
   (1 + a) + (1 + b) + (1 + c) + (1 + d) \geq 8
   \]

5. Equality in the AM-GM Inequality holds when \(1 + a = 1 + b = 1 + c = 1 + d\). Therefore, we set:
   \[
   1 + a = 1 + b = 1 + c = 1 + d = 2
   \]
   \[
   a = b = c = d = 1
   \]

6. Substituting \(a = b = c = d = 1\) into \(abcd\):
   \[
   abcd = 1 \cdot 1 \cdot 1 \cdot 1 = 1
   \]

Thus, the largest value of \(abcd\) is 1.

The final answer is \(\boxed{1}\)