Let $p_1,p_2, p_3,\dots,p_n$ be the first $n$ primes, and let $P_n$ be their product. Then the $p_{n+1}-2$ consecutive integers $P_n+2,P_n+3, \cdots, P_n+(p_{n+1}-1)$ are all composite.
For let $P_n+x$ be one of these numbers. Since $2\le x\lt p_{n+1}$, $x$ is divisible by some prime $p\le p_n$ ($x$ could itself be prime). But $P_n$ is also divisible by $p$, so $P_n+x$ is divisible by $p$. Clearly $P_n+x\gt p$, so $P_n+x$ is composite.
We can in general get a very slightly cheaper string by starting at $P_n-2$ and going backwards. These procedures get us arbitrarily long strings of consecutive composites, since there are infinitely many primes.
But one can do a lot better than $P_n$ in general. The subject of Prime Gaps has been extensively studied. You will find detailed information in this Wikipedia article.
In the paper Characterizing the Sum of Two Cubes, Kevin Broughan gives the relevant theorem,
Theorem: Let $N$ be a positive integer. Then the equation $N = x^3 + y^3$ has a solution in positive integers $x,y$ if and only if the following conditions are satisfied:
There exists a divisor $m|N$ with $N^{1/3}\leq m \leq (4N)^{1/3}.$
And $\sqrt{m^2-4\frac{m^2-N/m}{3}}$ is an integer.
The sequence of integers $F(n)$,
$$\begin{aligned}
F(n)
&= a^3+b^3 = (2 n + 6 n^2 + 6 n^3 + n^4)^3 + (n + 3 n^2 + 3 n^3 + 2 n^4)^3\\
&= c^3+d^3 = (1 + 4 n + 6 n^2 + 5 n^3 + 2 n^4)^3 + (-1 - 4 n - 6 n^2 - 2 n^3 + n^4)^3
\end{aligned}$$
for integer $n>3$ apparently is expressible as a sum of two positive integer cubes in exactly and only two ways.
$$\begin{aligned}
F(4) &= 744^3+756^3 = 945^3+15^3\\
F(5) &= 1535^3+1705^3 = 2046^3+204^3\\
&\;\vdots\\
F(60) &=14277720^3+26578860^3 = 27021841^3+12506159^3
\end{aligned}$$
Using Broughan's theorem, I have tested $F(n)$ from $n=4-60$ and, per $n$, it has only two solutions $m$, implying in that range it is a sum of two cubes in only two ways. Can somebody with a faster computer and better code test it for a higher range and see when (if ever) the proposed statement breaks down? Incidentally, we have the nice relations,
$$a+b = 3n(n+1)^3$$
$$c+d = 3n^3(n+1)$$
Note: $F(60)$ is already much beyond the range of taxicab $T_3$ which is the smallest number that is the sum of two positive integer cubes in three ways.
$$T_3 \approx 444.01^3 = 167^3+436^3 = 228^3+423^3 = 255^3+414^3$$
(Using the theorem, this yields 3 values for $m$.)
Best Answer
Consider the three odd composites $9,25,35$. These are, respectively, $0,1,2\pmod 3$. Thus, if $n$ is an even number, one of $n-9,n-25,n-35$ is an odd composite divisible by $3$ (well, supposing it is $>3$ at least). Thus $35+3=\fbox {38}$ is the largest even number that might be an example...inspection shows that it is indeed an example, hence the maximum example.