Solution:
It is easy to see that such a number can never be divisible by $2,3,5$, or $7$. They can be divisible by $11$, the smallest example being $4242424241 = 11 \cdot 547 \cdot 705073$.

What makes this problem hard is finding the next prime that can divide such a number. Let $T_{n} = \sum_{i=0}^{n} 42 \cdot 10^{2i}$. Then the numbers $T_{n}$ modulo a prime $p$ will always be periodic, since $T_{n} = 100 T_{n-1} + 42$, so we just need to compute one period and see if it contains $\pm 1$.

Thus we find that modulo $13$ we get $3,4,0,3, \ldots$, modulo $17$ we get $8,9,7,11,3,2,4,0,8, \ldots$, modulo $19$ we get $4,5,10,16,8,6,15,3,0,4, \ldots$, and modulo $23$ we get $19,10,7,6,21,3,20,18,2,12,0,19, \ldots$, so none of these primes can ever divide $T_{n} \pm 1$.

But $424241 = 29 \cdot 14629$, so $29$ can also divide numbers of this form. Therefore the smallest composite number that can divide $T_{n} \pm 1$ for some $n$ is $\mathbf{319}$, and the smallest such $n$ is $83$.