The number of common terms in the two sequences: 15, 19, 23, 27,.... , 415 and 14, 19, 24, 29,... , 464 is

The first arithmetic progression (AP) is:
15, 19, 23, 27,..., 415
First term $a_1=15$, common difference $d_1=4$.

The second AP is:
14, 19, 24, 29,..., 464
First term $a_2=14$, common difference $d_2=5$.

The first common term of both sequences is $a=19$.

The common terms themselves form an AP whose common difference $d$ is the LCM of the common differences of the two individual APs:
$d=\text{LCM}(4,5)=20$.

The common terms must be less than or equal to the minimum of the two series' end values, which is $min(415,464)=415$.

Let the number of common terms be $n$. The $n$-th common term is:
$T_n=19+(n-1)\times 20\le 415$
$(n-1)\times 20\le 396$
$n-1\le 19.8$
$n\le 20.8$

Since $n$ must be an integer, the maximum value of $n$ is 20.