Let $ d $ be the common difference. Then \begin{align*} \frac{1}{a_n a_{n + 1}} &= \frac{1}{a_n (a_n + d)} \\ &= \frac{1}{d} \cdot \frac{d}{a_n (a_n + d)} \\ &= \frac{1}{d} \cdot \frac{(a_n + d) - a_n}{a_n (a_n + d)} \\ &= \frac{1}{d} \left( \frac{1}{a_n} - \frac{1}{a_n + d} \right) \\ &= \frac{1}{d} \left( \frac{1}{a_n} - \frac{1}{a_{n + 1}} \right).\end{align*}Thus, \begin{align*} \frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \dots + \frac{1}{a_{4000} a_{4001}} &= \frac{1}{d} \left( \frac{1}{a_1} - \frac{1}{a_2} \right) + \frac{1}{d} \left( \frac{1}{a_2} - \frac{1}{a_3} \right) + \dots + \frac{1}{d} \left( \frac{1}{a_{4000}} - \frac{1}{a_{4001}} \right) \\ &= \frac{1}{d} \left( \frac{1}{a_1} - \frac{1}{a_{4001}} \right) \\ &= \frac{1}{d} \cdot \frac{a_{4001} - a_1}{a_1 a_{4001}}.\end{align*}Since we have an arithmetic sequence, $ a_{4001} - a_1 = 4000d, $ so \[\frac{1}{d} \cdot \frac{a_{4001} - a_1}{a_1 a_{4001}} = \frac{4000}{a_1 a_{4001}} = 10.\]Hence, $ a_1 a_{4001} = \frac{4000}{10} = 400 $. Then \[|a_1 - a_{4001}|^2 = a_1^2 - 2a_1 a_{4001} + a_{4001}^2 = (a_1 + a_{4001})^2 - 4a_1 a_{4001} = 50^2 - 4 \cdot 400 = 900,\]so $ |a_1 - a_{4001}| = \boxed{30} $.