The numbers $p, q, r$, and $t$ satisfy $p<q<r<t$. When these numbers are paired, each pair has a different sum and the four largest sums are 19,22,25, and 28 . What is the the sum of the possible values for $p$ ?
Using the four numbers $p, q, r, t$, the possible sums of pairs are

$$
p+q, p+r, p+t, q+r, q+t, r+t
$$

Since $p<q<r<t$, then $p+q<p+r<p+t$.

Also, $p+t<q+t<r+t$, which gives $p+q<p+r<p+t<q+t<r+t$.

Now $p+r<q+r<r+t$, but the relative size of $p+t$ and $q+r$ is unknown at this time.

This means that the four largest sums are $p+r, p+t, q+t, r+t$, although we do not know the relative size of the first two of these.

Since the four largest sums are 19,22, 25, and 28 , then $r+t=28$ and $q+t=25$.

Also, $p+t$ and $q+r$ are 22 and 19 in some order.

Case 1: $r+t=28$ and $q+t=25$ and $q+r=22$ and $p+t=19$

Adding the first three of these equations, we obtain

$$
(r+t)+(q+t)+(q+r)=28+25+22
$$

which gives $2 q+2 r+2 t=75$ and so $q+r+t=\frac{75}{2}$.

From this, we obtain $t=(q+r+t)-(q+r)=\frac{75}{2}-22=\frac{31}{2}$, from which $p=19-t=19-\frac{31}{2}=\frac{7}{2}$. We can verify that when $q=\frac{19}{2}$ and $r=\frac{25}{2}$, the equations are satisfied.

Case 2: $r+t=28$ and $q+t=25$ and $q+r=19$ and $p+t=22$

Adding the first three of these equations, we obtain

$$
(r+t)+(q+t)+(q+r)=28+25+19
$$

which gives $2 q+2 r+2 t=72$ and so $q+r+t=36$.

From this, we obtain $t=(q+r+t)-(q+r)=36-19=17$, from which $p=22-t=5$.

We can verify that when $q=8$ and $r=11$, the equations are satisfied.

The sum of the possible values of $p$ is thus $\frac{7}{2}+5$ or $\frac{17}{2}$.

ANSWER: $\frac{17}{2}$