Find the sum of all positive integers $b < 1000$ such that the base-$b$ integer $36_{b}$ is a perfect square and the base-$b$ integer $27_{b}$ is a perfect cube.
The first step is to convert $36_{b}$ and $27_{b}$ into base-10 numbers. Then, we can write \[36_{b} = 3b + 6\] and \[27_{b} = 2b + 7\]. It should also be noted that $8 \leq b < 1000$.
Because there are less perfect cubes than perfect squares for the restriction we are given on $b$, it is best to list out all the perfect cubes. Since the maximum $b$ can be is 1000 and $2$ • $1000 + 7 = 2007$, we can list all the perfect cubes less than 2007.
Now, $2b + 7$ must be one of \[3^3, 4^3, ... , 12^3\]. However, $2b + 7$ will always be odd, so we can eliminate the cubes of the even numbers and change our list of potential cubes to \[3^3, 5^3, 7^3, 9^3\text{, and }11^3\].
Because $3b + 6$ is a perfect square and is clearly divisible by 3, it must be divisible by 9, so $b + 2$ is divisible by 3. Thus the cube, which is \[2b + 7 = 2(b + 2) + 3\], must also be divisible by 3. Therefore, the only cubes that $2b + 7$ could potentially be now are $3^3$ and $9^3$.
We need to test both of these cubes to make sure $3b + 6$ is a perfect square.
$\textbf{Case 1:}$ If we set \[3^3 = 2b + 7\]so \[b = 10\]. If we plug this value of b into $3b + 6$, the expression equals $36$, which is indeed a perfect square.
$\textbf{Case 2:}$ If we set \[9^3 = 2b + 7\]so \[b = 361\]. If we plug this value of b into $3b + 6$, the expression equals $1089$, which is $33^2$.
We have proven that both $b = 10$ and $b = 361$ are the only solutions, so \[10 + 361 = \boxed{371}\]