I will try and prove that $a_n = a_{n-1} + n$ isn't the only solution.
$a_n$ divides $a_{n-1}+n$, So we can take 'm' to be the quotient. Note that m is an integer.
This gives us $ma_n=a_{n-1}+n$
Putting n=2 we get, $ma_2 = a_1 + 2$
As $a_2>a_1$ and both of them being being integers, $a_2-a_1\ge1$
$a_2\ge a_1+1$
$a_2 +1\ge a_1+2$
$a_2 +1\ge ma_2$
$(m-1)a_2\le 1$
$m\le {a_2+1\over a_2}$, Note that $a_2\ge 1$.
Trying out any value of $a_2$, we get, $m\le 2$ and because $m$ is an integer, $m=1$ or $m=2$.
Which give us $a_n=a_{n-1}+n$ or $2{a_n}=a_{n-1}+n$
Solve the first equation by telescopy and I don't know how to solve the second equation.
For first equation you will get $a_n= a_1 -1 + {n^{2}+n \over 2}$.You can take $a_1$ to be any non-negative integer. Try solving equation 2.
For $P\in\mathbb{Z}[t]$ and $d\in\mathbb{Z}$, let $S(P,d)$ be the set of pairs $(x,y)$ of distinct integers such that $P(x)-P(y)=d$.
The goal is to find all pairs $(P,d)$ such that $S(P,d)$ is infinite.
I'll give a solution for the case $d\ne 0$, and a partial solution for the case $d=0$ . . .
Case $(1)$:$\;d\ne 0$.
Thus, let $d$ be a nonzero integer, and let $K$ be the set of integer divisors of $d$.
Claim:$\;S(P,d)$ is infinite if and only if $P=at+b$ for some integer $b$ and some $a\in K$.
First suppose $P=at+b$ for some integer $b$ and some $a\in K$.
Write $d=ak$, for some $k\in K$.
Then for $x,y\in\mathbb{Z}$, we have
\begin{align*}
&P(x)-P(y)=d
\qquad\qquad\qquad\qquad\qquad\qquad
\\[4pt]
\iff\;&(ax+b)-(ay+b)=d\\[4pt]
\iff\;&a(x-y)=d\\[4pt]
\iff\;&x-y=k\\[4pt]
\end{align*}
hence $(y+k,y)\in S(P,d)$ for all integers $y$, so $S(P,d)$ is infinite.
Conversely, suppose $S(P,d)$ is infinite.
Then for $x,y\in\mathbb{Z}$ with $x\ne y$,
\begin{align*}
&P(x)-P(y)=d
\qquad\qquad\qquad\qquad\qquad\qquad
\\[4pt]
\implies\;&(x-y){\,\mid\,}d\\[4pt]
\implies\;&x=y+k,\;\text{for some}\;k\in K\\[4pt]
\end{align*}
Since $S(P,d)$ is infinite and $K$ is finite, it follows that there is some $k\in K$ such that the equation
$$
P(y+k)-P(y)=d
\qquad\qquad\qquad\qquad
$$
holds for infinitely many integers $y$.
Then letting $f\in\mathbb{Z}[t]$ be given by $f(t)=P(t+k)-P(t)-d$, we get that $f$ has infinitely many zeros, hence $f$ must be identically $0$.
\begin{align*}
\text{Then}\;&f=0\\[4pt]
\implies\;&P(t+k)-P(t)-d=0\\[4pt]
\implies\;&P'(t+k)-P'(t)=0\\[4pt]
&\;\;\;\text{[where $P'\in\mathbb{Z}[t]$ is the formal derivative of $P$]}\\[4pt]
\implies\;&P'=a,\;\text{for some integer}\;a\\[4pt]
\implies\;&P=at+b,\;\text{for some integers}\;a,b\\[4pt]
\end{align*}
Then letting $x,y$ be distinct integers such that $P(x)-P(y)=d$, we get
\begin{align*}
&P(x)-P(y)=d\\[4pt]
\implies\;&(ax+b)-(ay+b)=d
\qquad\qquad\qquad\qquad\qquad
\\[4pt]
\implies\;&a(x-y)=d\\[4pt]
\implies\;&a\in K\\[4pt]
\end{align*}
so $P=at+b$ for some integer $b$ and some $a\in K$, as claimed.
This completes the analysis for case $(1)$.
Case $(2)$:$\;d=0$.
For this case, I'll identify a class of polynomials $f\in\mathbb{Z}[t]$ such that $S(f,0)$ is infinite, and I'll conjecture that there are no others.
Let $V$ be the set of all polynomials $f\in\mathbb{Z}[t]$ such that $S(f,0)$ is infinite.
Some properties of the set $V$ . . .
Property $(0)$:$\;$If $f\in\mathbb{Z}[t]$ and $\deg(f)$ is odd, then $f\not\in V$.
Proof:$\;$Since $\deg(f)$ is odd, there exists $\theta\in (0,\infty)$ such that if $x,y\in\mathbb{R}$ with $x\ne y$ satisfy the equation $f(x)=f(y)$, then $x,y\in (-\theta,\theta)$.
Thus $(x,y)\in S(f,0)$ implies $x,y\in (-\theta,\theta)$, hence $S(f,0)$ is finite, so $f\not\in V$.
Property $(1)$:$\;t^2\in V$.
Proof:$\;$For all nonzero integers $y$, we have $(-y,y)\in S(t^2,0)$, hence $S(t^2,0)$ is infinite, so $t^2\in V$.
Property $(2)$:$\;t^2+t\in V$.
Proof:$\;$Let $f=t^2+t$.
Let $y\in\mathbb{Z}$ and let $x=-y-1$.
Then we have $x\ne y$ and
$$f(x)=x^2+x=(-y-1)^2+(-y-1)=(y^2+2y+1)-(y+1)=y^2+y=f(y)$$
thus for all $y\in\mathbb{Z}$, we have $(-y-1,y)\in S(t^2+t,0)$, hence $S(f,0)$ is infinite, so $t^2+t\in V$.
Property $(3)$:$\;$If $f\in V$, then $f(t+a)\in V$ for all $a\in\mathbb{Z}$.
Proof:$\;$Let $a\in\mathbb{Z}$ and let $g=f(t+a)$.
Then $(x,y)\in S(f,0)$ implies $(x-a,y-a)\in S(g,0)$, hence $S(g,0)$ is infinite, so $g\in V$.
Property $(4)$:$\;$If $f\in V$ and $g\in\mathbb{Z}[t]$, then $g\circ f\in V$.
Proof:$\;$Let $(x,y)\in S(f,0)$.
\begin{align*}
\text{Then}\;\;&(x,y)\in S(f,0)
\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\\[4pt]
\implies\;&f(x)=f(y)\\[4pt]
\implies\;&g(f(x))=g(f(y))\\[4pt]
\implies\;&(g\circ f)(x)=(g\circ f)(y)\\[4pt]
\implies\;&(x,y)\in S(g\circ f,0)\\[4pt]
\end{align*}
hence $S(g\circ f,0)$ is infinite, so $g\circ f\in V$.
Combining properties $(1),(2),(3),(4)$, we get
Property $(*)$:$\;$If $f=g\bigl((t+a)^2\bigr)$ or $f=g\bigl((t+a)^2+(t+a)\bigr)$ where $g\in\mathbb{Z}[t]$ and $a\in\mathbb{Z}$, then $f\in V$.
What about the converse of property $(*)$?
Conjecture:$\;$If $f\in V$, then for some $g\in\mathbb{Z}[t]$ and some $a\in\mathbb{Z}$, either $f=g\bigl((t+a)^2\bigr)$ or $f=g\bigl((t+a)^2+(t+a)\bigr)$.
Best Answer
Nate's hint (that $n^4 \equiv 0,1 \mod 16$ depending on whether $n$ is even or odd, respectively) gives you the answer. Since $65536\equiv 0 \mod 16$, if any of the $n_i$ are odd, then all of the $n_i$ must be odd so that their sum is $16\equiv 0 \mod 16$. But there are no two consecutive odd numbers. If you then look at all $n_i$ being even, you have the same problem. So there is no solution containing two consecutive values of $n_i$.