In a sequence of natural numbers $ a_1 $, $ a_2 $, $ \dots $, $ a_ {1999} $, $ a_n-a_ {n-1} -a_ {n-2} $ is divisible by $ 100 (3 \leq n \leq 1999) $. It is known that $ a_1 =  19$ and $ a_2 =  99$. Find the remainder of $ a_1 ^ 2 + a_2 ^ 2 + \dots + a_ {1999} ^ 2 $ by $8$.
1. Given the sequence \( a_1, a_2, \dots, a_{1999} \) where \( a_n - a_{n-1} - a_{n-2} \) is divisible by \( 100 \) for \( 3 \leq n \leq 1999 \), we start by noting that this implies:
   \[
   a_n \equiv a_{n-1} + a_{n-2} \pmod{100}
   \]
   We are given \( a_1 = 19 \) and \( a_2 = 99 \).

2. To find the remainder of \( a_1^2 + a_2^2 + \dots + a_{1999}^2 \) modulo \( 8 \), we first reduce the sequence modulo \( 4 \):
   \[
   a_1 \equiv 19 \equiv 3 \pmod{4}
   \]
   \[
   a_2 \equiv 99 \equiv 3 \pmod{4}
   \]
   Using the recurrence relation \( a_n \equiv a_{n-1} + a_{n-2} \pmod{4} \), we compute the next few terms:
   \[
   a_3 \equiv a_2 + a_1 \equiv 3 + 3 \equiv 6 \equiv 2 \pmod{4}
   \]
   \[
   a_4 \equiv a_3 + a_2 \equiv 2 + 3 \equiv 5 \equiv 1 \pmod{4}
   \]
   \[
   a_5 \equiv a_4 + a_3 \equiv 1 + 2 \equiv 3 \pmod{4}
   \]
   \[
   a_6 \equiv a_5 + a_4 \equiv 3 + 1 \equiv 4 \equiv 0 \pmod{4}
   \]
   \[
   a_7 \equiv a_6 + a_5 \equiv 0 + 3 \equiv 3 \pmod{4}
   \]
   We observe that the sequence is periodic with a period of 6: \( 3, 3, 2, 1, 3, 0 \).

3. Next, we calculate the sum of the squares of one period modulo \( 8 \):
   \[
   3^2 + 3^2 + 2^2 + 1^2 + 3^2 + 0^2 = 9 + 9 + 4 + 1 + 9 + 0 = 32 \equiv 0 \pmod{8}
   \]

4. Since the sequence is periodic with a period of 6, we need to determine how many complete periods fit into 1999 terms and the remainder:
   \[
   1999 \div 6 = 333 \text{ complete periods with a remainder of } 1
   \]
   Therefore, the sequence up to \( a_{1999} \) consists of 333 complete periods plus one additional term.

5. The sum of the squares of the complete periods is:
   \[
   333 \times 0 \equiv 0 \pmod{8}
   \]
   The additional term is \( a_1^2 \equiv 3^2 \equiv 9 \equiv 1 \pmod{8} \).

6. Adding the contributions from the complete periods and the additional term:
   \[
   0 + 1 \equiv 1 \pmod{8}
   \]

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