Given that $a_1, a_2, a_3, . . . , a_{99}$ is a permutation of $1, 2, 3, . . . , 99,$ find the maximum possible value of
$$|a_1 - 1| + |a_2 - 2| + |a_3 - 3| + \dots + |a_{99} - 99|.$$

1. We are given a permutation \(a_1, a_2, a_3, \ldots, a_{99}\) of the numbers \(1, 2, 3, \ldots, 99\). We need to find the maximum possible value of the expression:
   \[
   |a_1 - 1| + |a_2 - 2| + |a_3 - 3| + \dots + |a_{99} - 99|.
   \]

2. To maximize the sum of absolute differences, we should pair each \(a_i\) with \(i\) such that the difference \(|a_i - i|\) is as large as possible. This can be achieved by reversing the order of the permutation, i.e., setting \(a_i = 100 - i\).

3. Let's verify this by calculating the sum:
   \[
   |a_1 - 1| + |a_2 - 2| + |a_3 - 3| + \dots + |a_{99} - 99|
   \]
   where \(a_i = 100 - i\).

4. Substituting \(a_i = 100 - i\) into the expression, we get:
   \[
   |(100 - 1) - 1| + |(100 - 2) - 2| + |(100 - 3) - 3| + \dots + |(100 - 99) - 99|
   \]
   Simplifying each term, we have:
   \[
   |99 - 1| + |98 - 2| + |97 - 3| + \dots + |1 - 99|
   \]
   which simplifies to:
   \[
   98 + 96 + 94 + \dots + 2
   \]

5. Notice that this is an arithmetic series with the first term \(a = 98\), the common difference \(d = -2\), and the number of terms \(n = 49\) (since the sequence goes from 98 to 2).

6. The sum of an arithmetic series is given by:
   \[
   S_n = \frac{n}{2} (a + l)
   \]
   where \(l\) is the last term. Here, \(a = 98\), \(l = 2\), and \(n = 49\).

7. Substituting these values, we get:
   \[
   S_{49} = \frac{49}{2} (98 + 2) = \frac{49}{2} \cdot 100 = 49 \cdot 50 = 2450
   \]

8. Since the sequence is symmetric, the sum of the first 49 terms is equal to the sum of the last 49 terms. Therefore, the total sum is:
   \[
   2 \cdot 2450 = 4900
   \]

The final answer is \(\boxed{4900}\).