We are given that $$\frac{p+q}{2} \mid 100p+q, \frac{p+q}{2} \mid 100p+q.$$It follows that $$p+q \mid 200p+2q, p+q \mid 200q+2p.$$Clearly, $\gcd(p+q,p) = \gcd(p+q,q)=1,$ so we must have $$p+q \mid 198.$$We take cases on what $p+q$ is: p+q=198: Since $p$ and $q$ are distinct two digit primes, their maximal sum is $97+89=186<198,$ so this case yields no solutions. $\blacksquare$ p+q=99: Since $p+q$ is odd, one of them must be even, so one of $p,q$ is $2,$ but they must both have two digits, so this case yields no solutions. $\blacksquare$ p+q=66: Simply subtracting all two-digit primes less than $66$ and checking if the result is prime, we get the solutions $$\boxed{ \{13,53\}, \{19,47\}, \{23,43\}, \{29,37\}}.$$$\blacksquare$ p+q=33: Since $p+q$ is odd, one of them must be even, so one of $p,q$ is $2,$ but they must both have two digits, so this case yields no solutions. $\blacksquare$ p+q<=22: Since $p$ and $q$ are two-digit primes, their minimal sum is $11+13=24>22,$ so this case yields no solutions. $\blacksquare$ Hence, all possible pairs $\{p,q\}$ are $$\boxed{ \{13,53\}, \{19,47\}, \{23,43\}, \{29,37\}}.$$$\square$