Let $n = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$. Then $8 = (\alpha_1 + 1) \cdots (\alpha_k + 1)$, and $3240 = \frac{p_1^{\alpha_1+1}-1}{p_1-1} \cdots \frac{p_k^{\alpha_k+1}-1}{p_k-1}$. Hence there are three cases:
(a) $8 = \alpha_1 + 1 \implies n = p^7$.
(b) $8 = (\alpha_1 + 1)(\alpha_2 + 1) \implies n = p_1 p_2^3$.
(c) $8 = (\alpha_1 + 1)(\alpha_2 + 1)(\alpha_3 + 1) \implies n = p_1 p_2 p_3$.
Then we check $\sigma(n)$:
(a) $1 + p^2 + \cdots + p^7 = 3240$. We have $2 < p < 5$, so $p = 3$. But substituting yields no solution.
(b) $(p_1 + 1)(p_2^3 + p_2^2 + p_2 + 1) = 3240 \iff (p_1 + 1)(p_2 + 1)(p_2^2 + 1) = 3240$. First note that if $q$ is an odd prime then, by Fermat's theorem, $p_2^2 + 1 \equiv 0 \pmod{q} \implies 1 \equiv p_2^{q-1} \equiv (-1)^{\frac{q-1}{2}} \pmod{q} \implies q \equiv 1 \pmod{q}$. Since $3240 = 2^3 \cdot 3^4 \cdot 5$, the only primes that can divide $p_2^2 + 1$ are $2$ and $5$. This leaves the possibilities $p_2 = 2$ and $p_2 = 3$, none of which yield a solution.

(c) $(p_1+1)(p_2+1)(p_3+1) = 3240$. Let's consider some cases.
(c.1) One of the primes $p_i$ is $2$. Suppose that $p_1 = 2$. Then $(p_2+1)(p_3+1) = 1080 \iff (\frac{p_2+1}{2})(\frac{p_3+1}{2}) = 270$.
Let $x = \frac{p_2+1}{2}$ and $y = \frac{p_3+1}{2}$. Then $xy = 270$ is fixed and we want to minimize $2p_2p_3 = 2(2x-1)(2y-1) = 8 \cdot 270 - 4(x+y) + 2$, that is, we want to maximize $x+y$. This happens when $|x-y|$ is maximum. Since $p_2$ and $p_3$ are primes, the optimal values for $x$ and $y$ are $x=2$ and $y=135$, that is, $p_2 = 3$ and $p_3 = 269$, leading to the minimal solution $n = 1614$.
(c.2) All primes $p_i$ are odd. Then $(\frac{p_1+1}{2})(\frac{p_2+1}{2})(\frac{p_3+1}{2}) = 405 = 3^3 \cdot 5$. Then one prime, say $p_1$, is equal to $2 \cdot 3^k \cdot 5 - 1 = 10 \cdot 3^k - 1$; and since $\frac{p_i+1}{2}$ cannot be equal to $1$, then they must have at least one factor $3$ for $i=2,3$; there are only three factors $3$, so $k=0$ or $k=1$. $k=0$ is not possible; $k=1$ yields $p_1 = 29$ and $(\frac{p_2+1}{2})(\frac{p_3+1}{2}) = 27 \implies p_2 = 5$ and $p_3 = 17$, leading to $n = 2465$.
Hence the smallest value for $n$ is $1614$.