We begin by factoring the given integer into prime factors. Since 636405 ends in a 5, it is divisible by 5, so $636405=5 \times 127281$. Since the sum of the digits of 127281 is a multiple of 3, then it is a multiple of 3, so $636405=5 \times 3 \times 42427$. The new quotient (42427) is divisible by 7, which gives $636405=5 \times 3 \times 7 \times 6061$. We can proceed by systematic trial and error to see if 6061 is divisible by $11,13,17,19$, and so on. After some work, we can see that $6061=11 \times 551=11 \times 19 \times 29$. Therefore, $636405=3 \times 5 \times 7 \times 11 \times 19 \times 29$. We want to rewrite this as the product of three 2-digit numbers. Since $3 \times 5 \times 7=105$ which has three digits, and the product of any three of the six prime factors of 636405 is at least as large as this, then we cannot take the product of three of these prime factors to form a two-digit number. Thus, we have to combine the six prime factors in pairs. The prime factor 29 cannot be multiplied by any prime factor larger than 3, since $29 \times 3=87$ which has two digits, but $29 \times 5=145$, which has too many digits. This gives us $636405=87 \times 5 \times 7 \times 11 \times 19$. The prime factor 19 can be multiplied by 5 (since $19 \times 5=95$ which has two digits) but cannot be multiplied by any prime factor larger than 5, since $19 \times 7=133$, which has too many digits. This gives us $636405=87 \times 95 \times 7 \times 11=87 \times 95 \times 77$. The sum of these three 2-digit divisors is $87+95+77=259$.