How many terms are identical in the two APs 19, 21, 23, up to 110...

For the first AP: $19, 21, 23, \ldots$
First term $a_1 = 19$, common difference $d_1 = 2$, number of terms $n_1 = 110$.
Last term $L_1 = a_1 + (n_1 - 1)d_1 = 19 + (110 - 1) \times 2 = 19 + 218 = 237$.

For the second AP: $19, 22, 25, 28, \ldots$
First term $a_2 = 19$, common difference $d_2 = 3$, number of terms $n_2 = 75$.
Last term $L_2 = a_2 + (n_2 - 1)d_2 = 19 + (75 - 1) \times 3 = 19 + 222 = 241$.

The identical terms of both APs form a new AP.
The first common term is $a = 19$.
The common difference of the new AP is $D = \text{LCM}(d_1, d_2) = \text{LCM}(2,3) = 6$.

The last term of this new AP must be less than or equal to the smaller of the two last terms, which is $\min(237,241) = 237$.
Let the number of identical terms be $n$.

$a + (n - 1)D \le 237$

$19 + (n - 1) \times 6 \le 237$

$6(n - 1) \le 218$

$n - 1 \le \dfrac{218}{6}$

$n - 1 \le 36.33$

$n \le 37.33$

Since $n$ must be an integer, the maximum possible value for $n$ is $37$.

Answer: $37$