Three of them

Solving for $x$ and $y$ will give:

$x=\dfrac{56-3p}5 \text{ and } y= \dfrac{4p-13}5....(1)$ $\because x,y>0\implies \dfrac{13}{4}< p<\dfrac{56}3....(2)$

Since $p\in\mathrm{Z}$ $\implies p\in\{4,5,6,...,18\}$ .

From $1$ , now since $y$ is an integer it's numerator must be a multiple of $5$ and for that $4p$ must end in $8$ or $3$ so when $13$ is subtracted, the numerator ends in $5$ or $0$ respectively and hence divisible by $5$ . $4p$ cannot end in $3$ but ends in $8$ when $p\equiv 5n+2$ or $p\in\{7,12,17\}$ . For these values of $p,$ $x$ also assumes an integer.

$p=7,~~ (x,y)=(7,3)$ $p=12, ~~(x,y)=(4,7)$ $p=17,~~(x,y)=(1,11)$ Summing them as per question gives $\boxed{69}$ .

Graphing X+2Y=6+p, .. and..2X-Y=25-2p for p= 1 to 18 when we encounter negative values.