jmc

$4641$ has a prime factorization of $4641=3\cdot 7\cdot 13\cdot 17$. Multiplying any $3$ of its prime factors yields a number that is greater than $100$, so the two $2$-digit numbers must each be the product of $2$ of its prime factors. The only other prime factor $17$ can be multiplied by without getting a $3$-digit number is $3$, so one of them must be $17\cdot 3=51$. That makes the other $7\cdot 13=91$. $51+91=\boxed{142}$.