Here's my brute force example... based on convergence of 1/n versus divergence of 1/n^2.
The fact that $\sum \frac{1}{n}$ diverges, while $\sum \frac{1}{n^2}$ converges, can be used to construct strictly increasing integer sequences $a_n$ and $b_n$ for which each of $\sum \frac{1}{a_n}$ and $\sum \frac{1}{b_n}$ diverges, but nevertheless $\sum \frac{1}{max(a_n,b_n)}$ converges. To describe these sequences, let the number $k=1.5$ be called the "divergence bound". What we do is alternately put sequential numbers in one of the sequences, while putting squares in the other, making sure to put enough sequential numbers in each block in each case to exceed the divergence bound. Then a sequence of adjacent squares is placed in the other sequence, after which we switch the roles of the two sequences, and iterate. We make sure that in each case the squared number exceeds the corresponding number in the sequence going sequentially, so that when the max is taken one gets only squares; we also take care that we do not repeat squares, so that the resulting sequence of max terms is a subsequence of the sequence $1^2,2^2,3^2,...$, which ends up making the sum of max reciprocals converge.
I'll call $A$ the list of $a_k$ and $B$ the list of $b_k$. We will begin $A$ with a sequential part; since $1/1+1/2=1.5$ we may begin $a$ by $[1,2]$. For corresponding squares in $B$ we begin $B$ by $[1^2,2^2]$. We now want to switch to making adjacent integers in $B$. So we choose the least integer exceeding the present last term of $B$, which is 5. We find that $1/5+1/6+...+1/20$ is about 1.51 (stopping at 1/19 was too small]. So now we have $B=[1^2,2^2,5,6,...,20]$ The first square we haven't yet used is $3^2$, and (luckily) $3^2>5$, so we may now put $3^2$ into $A$ in its third position, and continue with $4^2,5^2,...,18^2$. At this point we have "done two blocks" of each sequence for $A$ and $B$, so that the beginnings of these are $A=[1,2,3^2,4^2,...,18^2],$ $B=[1^2,2^2,5,6,7,...,20].$ Now $18^2=324$, so the next entry of $A$ should be 325, in order to maintain an increasing sequence. And we may begin the squares in the next part of $B$ with $19^2$, since it happens to exceed 325. (In such cases, if the next square not yet used is not large enough, we may simply choose a larger square). For $A$ our next sequence is to be adjacent, and achieve the divergence bound of 1.5; I found we had to go all the way from 325 to 1453 for this. So the next block of adjacent $A$ values is quite long. The corresponding squares to be placed under these in $B$ may be taken to be $19^2,20^2,...,1147^2$.
Note: at the junctures I chose a square to place in one of the sequences, at the beginning of the part of the other which was to run in adjacent increasing form. The first square was say $a^2$ and at that term in the other was say $b$. I did make sure that $a^2>b$, but didn't explicitly say that the sequence of squares starting at $a^2$ automatically should exceed term by term the numbers following $b$. However this is clear, since the differences between succesive squares is always greater than 1, while the other sequence is only going up 1 each time.
It seems clear this construction will give what we want, since the partial sums of reciprocals for both of $A,B$ will exceed arbitrary multiples of the divergence bound of 1.5, making these series diverge, and yet the series of reciprocal max terms is bounded above by the sum of reciprocals of squares.