The answer is $5$, achieved for example with the integers $575$ to $579$.

Let $d(n)$ denote the sum of the divisors of $n$. I claim that if $n$ is even, then $n$ can only be resistant if it is either a square or twice a square. This holds since if we have a prime $p>2$ such that $v_p(n)=k$ where $k$ is odd, then the sum of divisors of $p^k$ is $d(p^k)=p^k+p^{k-1}+\dots+1$, an even integer. Since $d(n)$ is multiplicative, $d(n)=d(p^k)\cdot d\left(\frac{n}{p^k}\right)$, thus $d(n)$ is divisible by $2$, but so is $n$, so $gcd(n,d(n))\geq 2$ and thus $n$ is not resistant. Now note that if we have 6 consecutive resistant integers, then 3 of them are even and must be of the form $x^2$ or $2y^2$. It is also clear that the difference between any two consecutive even square numbers is at least 12, and the difference between any 2 numbers of the form $2y^2$ is at least 6. Thus if the 3 resistant even integers are $2k$, $2k+2$ and $2k+4$, then either $2k=x^2$, $2k+2=2y^2$ and $2k+4=z^2$ for some $x,y,z$, or $2k=2x^2$, $2k+2=y^2$, $2k+4=2z^2$ for some $x,y,z$. In both cases $2k$ and $2k+4$ are both of the same form (either $x^2$ or $2y^2$), but differ by less than 6, contradiction. This proves that the answer is $\leq 5$.