Express the numbers $n-1,n,n+1$ in their prime factorization. Let $x$ be any one of $n-1,n,n+1$ and express \[x=2^a.3^b.5^c\cdots.\]Then \[s(x)=(1+2+…2^a).(1+3+…3b).(1+5+…5c)...\] Assume now that all of $s(n-1),s(n),s(n+1)$ are odd The first term is always odd, while the remaining terms are only odd if the exponents $b,c,…$ are even, so $x$ must be a multiple of a power of $2$ and an odd perfect square. Now if $n-1,n+1$ are odd, they must both be odd perfect squares. But this is impossible as no two positive perfect squares differ by $2$. Therefore, $n$ is odd, and hence an odd perfect square. Express $n=\alpha ^2$ where $\alpha=2k+1,k\in \mathbb Z^{+}.$ Then we have \[n-1=\alpha^2-1=(2k+1)^2-1=4k(k+1)\]\[n=\alpha^2=(2k+1)^2=4k^2+4k+1\]\[n+1=\alpha^2+1=(2k+1)^2=2(2k^2+2k+1)\]$2k^2+2k+1$ is odd, so must be an odd perfect square. So $n+1$ is twice an odd perfect square. One of $k,k+1$ must be odd. Also, gcd$(k,k+1)=1.$ So any odd prime factors in $k$ are not in $k+1,$ and vice-versa. Let $y$ be the even number in {$k,k+1$} and $z$ be the other. Note that $y$ must be of the form $2^q.r^2$ where $r$ is an odd number, and $z$ must be an odd perfect square, of the form $s^2$ where $s$ is odd. If $q$ is even, then $y$ is a perfect square, which is impossible because $y$ and $z$ would both then be positive perfect squares differing by $1.$ Therefore, we have shown that $q$ is odd, and so \[4k(k+1)=4yz=4.2^q.r^2.s^2=2.2^{q+1}.r^2.s^2=2(2^{\frac{q+1}{2}}rs)^2.\] Since $q$ is odd, $\frac{q+1}{2}$ is an integer, and hence we can conclude that $n-1=4k(k+1)$ is twice a square number. To recap, we now have that $n-1$ is twice a square number, and $n+1$ is twice an odd perfect square, with $n$ being odd. Therefore, $n-12$ and $n+12$ are both positive perfect squares. However, this is impossible because they differ by one. We have arrived at a fatal contradiction of the original assumption, so $s(n-1),s(n),s(n+1)$ cannot all be odd.

Note that $n=\prod p_i^{a_i}$ then $s(n)=\prod (1+p_i+...+p_i^{a_i})$. Now $s(n)$ is odd iff $n=2^xm^2$ , for$ x $nonegative and $m$ odd. Assume $s(n-1)s(n)s(n+1)$ is odd. So we have first case if $n$ is even: $n+1=x^2$ and $n-1=y^2$ for some $x,y$ which is contradiction because $x^2-y^2\equiv 0,1-1(mod 4)$. Second case if $n$ is odd : $n+1=2^xp^2$ ,$n-1=2^yq^2$, $n=k^2$ , for positive $x,y$ and odd $p,q,k$ and we have that one of $x,y$ must be $1$: If $y=1$ by $(mod 8)$ $2^xp^2\equiv 4(mod 8)$ so $x=2$ but $4p^2=k^2+1$ gives contradiction. If $x=1$ and we have $n+1=2p^2$ and $n-1=2(2^mq)^2$ for some $m$ positive (since $n=k^2$ power of two dividing $n-1$ is odd), so $ 2(p^2-(2^mq)^2)=2 $ which is contradiction.