This post was inspired by the following one.
Fast way to check if two integers don't have any prime factors in common
We consider the sequence of numbers of equal $(p+q)$ for a number $N=pq$. We consider only odd composite numbers. We consider the example of $N=5*7=35=1*35$. The sequence of numbers of equal $(p+q)=1+35=36$ is given by $$35,68,99,128,155,180,203,224,243,260,275,288,299,308,315,320,323,324$$
There is another sequence of numbers of equal $(p+q)=5+7$ for $N=35$ but in this case we do not know the other numbers with the same $(p+q)$ unless we factor $N=35$. So we do not consider this sequence.
In the first sequence given above, we know that there are numbers with a common factor with $N=35$ every time we hit a multiple of $5$ or $7$. To find them, we need to take the $GCD$ of $N$ and those numbers with a common factor. Because we don't know which number has a common factor with $N$, we need to take the $GCD$ with $N=35$ of every number. This method cannot be efficient. However, I noticed that there are numbers with a common factor with $N$ but also with the same sum of digits ($SD=8$). For the example above these numbers are given by $$224,260,323$$
So we only need to check $3$ numbers out of $17$. Of these $3$ numbers only $224$ and $260$ have a common factor with $N=35$.
When experimenting with the method, I always found numbers with the same sum of digits and also found at least one with a common factor.
We know that the sum of digit is periodic within the sequence of numbers with the same $(p+q)$ so we only need to calculate those numbers and check if they have a common factor with $N$.
The question is this: Can it be proven that there is always at least one number (with the same sum of digits as $N$) within the sequence with a common factor with $N$?