Let $a_1,a_2,\dots \dots .a_n$ be n numbers each of which is either $1 \text{or }-1.$ If $a_1{a}_2{a}_3{a}_4+a_2{a}_3{a}_4{a}_5+\dots \dots \dots +a_n{a}_1{a}_2{a}_3=0$, then

By multiplying all the terms $a_1^4{a}_2^4\dots a_n^4=1$

$(-1{)}^x\cdot 1^{n-x}=a_1^4{a}_2^4\dots a_n^4=1$$\Rightarrow x must be even. So 2/x$

$$\begin{gathered} {\mathrm{a}}_1{\mathrm{a}}_2{\mathrm{a}}_3{\mathrm{a}}_4+{\mathrm{a}}_2{\mathrm{a}}_3{\mathrm{a}}_4{\mathrm{a}}_5+\cdots +{\mathrm{a}}_{\mathrm{n}}{\mathrm{a}}_1{\mathrm{a}}_2{\mathrm{a}}_3=0 \\ \Rightarrow \mathrm{x}=\mathrm{n}-\mathrm{x} \\ \therefore \mathrm{n}=2\mathrm{x} \\ \mathrm{n} \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 4 \end{gathered}$$

or

Since each term in the sum is 1 or –1 and for sum to be 200, the number of terms must be even take n = 2m. Now there are exactly m products among $a_1{a}_2{a}_3{a}_4,a_2{a}_3{a}_4{a}_5,\dots \dots ,a_n{a}_1{a}_2{a}_3$ which are equal
to –1. Therefore there are m changes of sign in the sequence $a_1{a}_2,\dots \dots \dots ,a_n,a_1\Rightarrow m$ must be even

$\Rightarrow n$ must be divisible by 4