jmc

The sum of the divisors of $2^i{3}^j$ is equal to $(1+2^1+2^2+\cdots +2^{i-1}+2^i)(1+3^1+3^2+\cdots +3^{j-1}+3^j)=600,$ since each factor of $2^i{3}^j$ is represented exactly once in the sum that results when the product is expanded. Let $A=1+2^1+2^2+\cdots +2^i$ and $B=1+3^1+3^2+\cdots +3^j$, so that $A\times B=600$. The prime factorization of $600$ is $600=2^3\cdot 3\cdot 5^2$.

Notice that $A$ is the sum of $1$ and an even number and $B$ is the sum of $1$ and a number divisible by $3$. Thus, $A$ is odd and $B$ is not divisible by $3$. It follows that $A$ is divisible by $3$ and $B$ is divisible by $2^3$. We now have three separate cases: $(A,B)=(3\cdot 25,8),(3\cdot 5,8\cdot 5),(3,8\cdot 25)$.

In the first case, $B=1+3+\cdots +3^j=8$; for $j=1$, we have that $1+3=4<8$, and for $j=2$, we have that $1+3+9=13>8$. Thus, this case is not possible.

For the third case, $B=1+3+\cdots +3^j=200$; for $j=4$, then $1+3+9+27+81=121<200$, and for $j=5$, we have that $1+3+9+27+81+243=364>200$. Thus, this case is also not possible.

It follows that $(A,B)=(15,40)$, in which case we find that $i=j=3$ works. Thus, the answer is $3+3=\boxed{6}$.