Nine positive integers $a_1,a_2,...,a_9$ have their last $2$-digit part equal to $11,12,13,14,15,16,17,18$ and $19$ respectively. Find the last $2$-digit part of the sum of their squares.
1. We need to find the last two digits of the sum of the squares of the integers \( a_1, a_2, \ldots, a_9 \) where the last two digits of these integers are \( 11, 12, 13, 14, 15, 16, 17, 18, \) and \( 19 \) respectively. This means we are interested in the sum modulo 100.

2. First, we calculate the squares of each of these numbers and then take each result modulo 100:
   \[
   \begin{aligned}
   11^2 & = 121 \equiv 21 \pmod{100}, \\
   12^2 & = 144 \equiv 44 \pmod{100}, \\
   13^2 & = 169 \equiv 69 \pmod{100}, \\
   14^2 & = 196 \equiv 96 \pmod{100}, \\
   15^2 & = 225 \equiv 25 \pmod{100}, \\
   16^2 & = 256 \equiv 56 \pmod{100}, \\
   17^2 & = 289 \equiv 89 \pmod{100}, \\
   18^2 & = 324 \equiv 24 \pmod{100}, \\
   19^2 & = 361 \equiv 61 \pmod{100}.
   \end{aligned}
   \]

3. Next, we sum these results:
   \[
   21 + 44 + 69 + 96 + 25 + 56 + 89 + 24 + 61.
   \]

4. We can add these numbers step-by-step to ensure accuracy:
   \[
   \begin{aligned}
   21 + 44 & = 65, \\
   65 + 69 & = 134 \equiv 34 \pmod{100}, \\
   34 + 96 & = 130 \equiv 30 \pmod{100}, \\
   30 + 25 & = 55, \\
   55 + 56 & = 111 \equiv 11 \pmod{100}, \\
   11 + 89 & = 100 \equiv 0 \pmod{100}, \\
   0 + 24 & = 24, \\
   24 + 61 & = 85.
   \end{aligned}
   \]

5. Therefore, the last two digits of the sum of the squares of the given integers are \( 85 \).

The final answer is \(\boxed{85}\).