How do I prove that for every positive integer $n$, there exist $n$ consecutive positive integers, each of which is composite?

To expand on the other solution already given,

Proof. Assume $m$ and $n$ exist in the positive integers and that $m$ is less than $n$. If $m$ is less than $n$, then $$ n!=1\cdot2\cdot3\cdot\dotsb\cdot m\cdot\dotsb\cdot n, $$

which is to say that $m$ is a factor of $n!$. So, \begin{align} m+n! &= m+(1\cdot2\cdot\dotsb\cdot m\cdot\dotsb\cdot n) \\ &= m\left(\frac{1\cdot 2\cdot \dotsb\cdot n}{m + 1}\right) \end{align} Remember that $m$ is a factor of $n!$ and so $n!/m$ is still an integer. So since $m$ is an integer that is bounded between $1$ and $n$, it stands that whatever number you pick up to $n$ can divide $m+n!$ making it composite till the $n$^th integer, but $n!$ has that $n$^th integer in it so the $n$^th integer is also composite which means that you can pick any integer between $1$ and $n$ inclusively and it will be composite.

The numbers you are given are consecutive, right? Are they composite? Hint: For the term $i + (n+1)!$, try to factor out $i$...

This is called spoonfeeding, and human students are notorious for needing it.

Does this mean that if $n = 5$, for example, then somewhere in the positive integers there are 5 consecutive composite integers, and that we want to prove that?

Yes, that's exactly what that means. For example, $24, 25, 26, 27, 28$ are five consecutive composite integers. But $29$ is prime, so if you want a run of six consecutive composite integers, you have to go up a little further to find $90, 91, 92, 93, 94, 95, 96$. Actually, that's seven consecutive composites.

Of course it's inefficient to run through the integers in the hopes of finding a run of composite numbers of the desired length. If only there was some formula that guaranteed us a run of at least $n$ consecutive composites, but unfortunately there isn't. What woe, to be in such a hopeless situation!

Oh, wait a minute, what is this?

Consider the numbers $$2 + (n + 1)!, 3 + (n + 1)!, \ldots, n + (n + 1)!, n + 1 + (n + 1)!$$

We know that $n!$ is even for $n > 1$, which then means that $2 + (n + 1)!$ must also be even. And we also know that $n!$ is divisible by $3$ for $n > 2$, so $3 + (n + 1)!$ must also be divisible by $3$. And we also know that $n!$ is divisible by $4$ for $n > 3$... see where I'm going with this?