Case 1 : $\rm n$ is odd. Set $\rm m=a\cdot ord_n(2016),a\equiv -(ord_n(2016))^{-1}\pmod n$. Then we have $\rm 2016^m+m=2016^{a\cdot ord_n(2016)}+a\cdot ord_n(2016)\equiv 1-(ord_n(2016))^{-1} \cdot ord_n(2016)\equiv 0\pmod n$ Case 2:$\rm n$ is even. Let $\rm n=2^qr,2\not |r$.Set $\rm m=2^qord_r(2016)f,f\equiv -(ord_r(2016))^{-1}(2^q)^{-1}\pmod r$. Then $\rm 2^q\mid 2016^{2^q}\mid 2016^m$ and $\rm 2^q\mid 2^q\mid 2^{2^q}\mid 2^m$ so $\rm 2^q\mid 2016^m+m$. Also $\rm 2016^m+m\equiv 2016^{2^qord_r(2016)f}+2^qord_r(2016)f\equiv 1-2^qord_r(2016)(ord_r(2016))^{-1}(2^q)^{-1}\equiv 1-1\equiv 0\pmod r$ So since $\rm (r,2^q)$ we have $\rm n\mid 2016^m+m$