We are told that $(x-a)(x-6)+3=(x+b)(x+c)$ for all real numbers $x$. In particular, this equation holds when $x=6$. Substituting $x=6$ gives $(6-a)(6-6)+3=(6+b)(6+c)$ or $3=(6+b)(6+c)$. Since $b$ and $c$ are integers, then $6+b$ and $6+c$ are integers, which means that $6+b$ is a divisor of 3. Therefore, the possible values of $6+b$ are $3,1,-1,-3$. These yield values for $b$ of $-3,-5,-7,-9$. We need to confirm that each of these values for $b$ gives integer values for $a$ and $c$. If $b=-3$, then $6+b=3$. The equation $3=(6+b)(6+c)$ tells us that $6+c=1$ and so $c=-5$. When $b=-3$ and $c=-5$, the original equation becomes $(x-a)(x-6)+3=(x-3)(x-5)$. Expanding the right side gives $(x-a)(x-6)+3=x^{2}-8x+15$ and so $(x-a)(x-6)=x^{2}-8x+12$. The quadratic $x^{2}-8x+12$ factors as $(x-2)(x-6)$ and so $a=2$ and this equation is an identity that is true for all real numbers $x$. Similarly, if $b=-5$, then $c=-3$ and $a=2$. (This is because $b$ and $c$ are interchangeable in the original equation.) Also, if $b=-7$, then $c=-9$ and we can check that $a=10$. Similarly, if $b=-9$, then $c=-7$ and $a=10$. Therefore, the possible values of $b$ are $b=-3,-5,-7,-9$. The sum of these values is $(-3)+(-5)+(-7)+(-9)=-24$.