, 2022

Solution 1

Since the sum of $p,q,r,s$ and the
sum of $q,r,s,t$ are both equal to
35, and $q,r,s$ are added in each
sum, then $p = t$.

Similarly, since the sum of $q,r,s,t$
and the sum of $r,s,t,u$ are both
equal to 35, and $r,s,t$ are added in
each sum, then $q=u$.

It can similarly be shown that $r=v$
and $s=w$.

Using the above observations, the sequence $p,q,r,s,t,u,v,w$ can be written as $p,q,r,s,p,q,r,s$.

Since the sum of $q$ and $v$ is 14 and $r=v$, then the sum of $q$ and $r$ is 14.

Since the sum of $p,q,r,s$ is 35 and
the sum of $q$ and $r$ is 14, then the sum of $p$ and $s$ is $35-14=21$.

The value of $p$ is as large as
possible exactly when the value of $s$ is as small as possible.

Since $s$ is a positive integer, its
smallest possible value is 1.

Therefore, the largest possible value of $p$ is $21-1=20$.

(We note that $20,4,10,1,20,4,10,1$
is an example of such a list.)

Solution 2

The sum of the values of each group of four consecutive letters is
35.

Thus, $p+q+r+s=35$ and $t+u+v+w=35$, and so $(p+q+r+s)+(t+u+v+w)=35+35=70$.

Rearranging the sum of these eight letters, we get $$p+q+r+s+t+u+v+w=(p+w)+(q+v)+(r+s+t+u)=70$$
However, $r+s+t+u=35$ (the sum of the
values of four consecutive letters), and $q+v=14$.

Substituting, we get $(p+w)+14+35=70$, and so $p+w=21$.

The value of $p$ is as large as
possible exactly when the value of $w$ is as small as possible.

Since $w$ is a positive integer, its
smallest possible value is 1.

Substituting, we get $p+1=21$, and so
the largest possible value of $p$ is
20.

(We note that $20,12,2,1,20,12,2,1$
is an example of such a list.)