let $x_ige 0(i=1,2,3,cdots,2009$,and such $$(x_1+x_2+cdots+x_2009)^2=4(x_1x_2+x_2x_3+cdots+x_2009x_1)=4$$
find the range $sum_i=1^2009x^2_i$

I guess the range is $[1.5,2]$
because I think when $x_1=1,x_2=1,x_3=x_4=cdots=x_2009=0$then is maximum of value $2$

and when $x_1=0.5,x_2=1,x_3=0.5,x_4=cdots=x_2009=0$ is minmum of the value $1.5$

I think you are right!

We'll prove that
$$(x_1+x_2+...+x_2009)^2geq4(x_1x_2+x_2x_3+...+x_2009x_1)$$ for all non-negatives $x_i$.

Indeed, let $x_2=maxlimits_ix_i$.

Thus, $$x_2x_2009geq x_1x_2009$$ and by AM-GM we obtain:
$$4(x_1x_2+x_2x_3+...+x_2009x_1)leq4(x_1+x_3+...+x_2009)(x_2+x_4+...+x_2008)leqleft(sumlimits_i=1^2009x_iright)^2,$$
where the equality occurs for
$$x_1(x_4+x_6+...+x_2008)=0,$$
$$x_2(x_5+...+x_2007)=0,$$ which gives
$$x_5=...=x_2007=0.$$
Now, if $x_1=0$ then $x_2x_3=1,$ which gives
$$2=sum_i=1^2009x_igeq x_2+x_3geq2sqrtx_2x_3=2,$$
which gives $x_2=x_3=1$, $x_1=x_4=...=x_2009=0$ and $sumlimits_i=1^2009x_i^2=2$.

If $x_1>0$ we obtain:
$$x_1(x_4+...+x_2008)=0,$$
which gives $x_4=x_5=...=x_2008=0.$

Now, let $x_2=a$, $x_1=b$,$x_3=c$ and $x_2009=d$.

Thus, $$(a+b+c+d)^2=4(ab+bd+ca)=4,$$
which gives $a+b+c+d=2$ and $ab+bd+ca=1.$

Thus, by AM-GM $$1=ab+bd+ca=ab+bd+dc+ca-dc=(a+d)(b+c)-dcleq$$
$$leqleft(fraca+b+c+d2right)^2-dc=1-dcleq1,$$
which gives $dc=0$ and $a+d=b+c=1.$

Now, let $c=0$.

Thus, $b=1=a+d$ and since $ageq b$, we obtain $d=0$ and $sumlimits_i=1^2009x_i^2=2$ again.

Let $d=0$.

Id est, by C-S
$$sum_i=1^2009x_i^2=a^2+b^2+c^2=1+frac12(1+1)(b^2+c^2)geq1+frac12(b+c)^2=1.5.$$
The equality occurs for $a=1$ and $b=c=frac12,$ which says that we got a minimal value.

Also, $$sum_i=1^2009x_i^2=a^2+b^2+c^2=1+(b+c)^2-2bcleq1+(b+c)^2=2.$$
The equality occurs for $a=b=1$ and $c=0,$ which says that we got a maximal value.

Solution

We are to state a lemma, on which the whole thing is based.

Lemma

Let $n > 4$ and $x_0,x_1,ldots,x_n,x_n+1$ be non-negative real
numbers, with $x_0=x_n$ and $x_n+1 = x_1$. Then $$left (sum_i=1^n x_i right )^2 geq 4sum_i=1^n
x_ix_i+1,$$
with the equality holding if and only if, for any $j~~ (1leq jleq n)$ we have$$x_j =dfrac 12 sumlimits_i=1^n x_i = x_j-1+x_j+1$$ and the rest
of the variables are equal to $0$.

Come back to the present case. Since $n=2009> 4$, $sumlimits_i=1^n x_i = 2$ and $sumlimits_i=1^n x_ix_i+1 = 1$, thus the equality holds. Therefore, for any $j~~~(1 leq j leq 2009)$, we have $x_j=1$,$x_j-1+x_j+1=1,$ and the rest are equal to $0$.

Under these constraints, we obtain $$S = sum_i=1^n x_i^2 = x_j^2 + x_j-1^2 + x_j+1^2 =1 + x_j-1^2 + x_j+1^2,$$ where $x_j-1, x_j+1 geq 0$ and $x_j-1+x_j+1= 1$. Thus, $dfrac 12 leq x_j-1^2 + x_j+1^2 leq 1$, therefore $$dfrac 32 leq S leq 2.$$