1. Let $f$ be a polynomial with positive integer coefficients, and let $n$ be a positive integer. We need to prove that $f(n)$ divides $f(f(n) + 1)$ if and only if $n = 1$.

2. First, consider the polynomial $f(x)$. Since $f$ has positive integer coefficients, $f(x)$ is strictly increasing for $x \geq 1$. This means that for any $x \geq 1$, $f(x) \geq f(1)$.

3. We need to show that $f(n) \mid f(f(n) + 1)$ if and only if $n = 1$.

4. Notice that $f(n) + 1 - 1 = f(n)$. Therefore, we can write:
$$f(n) \mid f(f(n) + 1) - f(1)$$
This follows from the fact that if $a \mid b$ and $a \mid c$, then $a \mid (b - c)$.

5. Since $f(n) \mid f(f(n) + 1) - f(1)$, it implies that:
$$f(n) \mid f(f(n) + 1)$$
if and only if:
$$f(n) \mid f(1)$$

6. Now, consider the case when $n = 1$:
$$f(1) \mid f(f(1) + 1)$$
Since $f$ has positive integer coefficients, $f(1)$ is a positive integer. Therefore, $f(1)$ divides $f(f(1) + 1)$.

7. Next, consider the case when $n > 1$. Since $f$ is strictly increasing and has positive integer coefficients, we have:
$$f(n) > f(1)$$
Therefore, $f(n)$ cannot divide $f(1)$ because $f(n)$ is greater than $f(1)$.

8. Hence, the only possible value for $n$ such that $f(n) \mid f(f(n) + 1)$ is $n = 1$.

$\blacksquare$