Problem 1853

We need to find all positive integer pairs $(a, n)$ such that $\frac{(a+1)^n - a^n}{n}$ is an integer.

First, observe that for $\frac{(a+1)^n - a^n}{n}$ to be an integer, $(a+1)^n - a^n$ must be divisible by $n$.

Consider the smallest prime divisor $p$ of $n$. We have:
$$(a+1)^n \equiv a^n \pmod{p}.$$
This implies:
$$\left(\frac{a+1}{a}\right)^n \equiv 1 \pmod{p}.$$
Let $z \equiv \frac{a+1}{a} \pmod{p}$. Then:
$$z^n \equiv 1 \pmod{p}.$$
By Fermat's Little Theorem, we know:
$$z^{p-1} \equiv 1 \pmod{p}.$$
Since $z^n \equiv 1 \pmod{p}$, it follows that:
$$z^{\gcd(n, p-1)} \equiv 1 \pmod{p}.$$
Given that $p$ is the smallest prime divisor of $n$, we have $\gcd(n, p-1) = 1$. Therefore:
$$z \equiv 1 \pmod{p}.$$
This implies:
$$\frac{a+1}{a} \equiv 1 \pmod{p},$$
which simplifies to:
$$a+1 \equiv a \pmod{p}.$$
Thus:
$$1 \equiv 0 \pmod{p},$$
which is a contradiction unless $n = 1$.

Therefore, the only solution is when $n = 1$. In this case, $\frac{(a+1)^1 - a^1}{1} = 1$, which is always an integer for any positive integer $a$.

Hence, the solution is:
$$\boxed{(a, n) = (a, 1)}$$
for any positive integer $a$.