In order to study the existence of positive periodic solutions, we first introduce the continuation theorem as follows.

Our main result is given in the following theorem.

Proof.
Set N(t)=(N1(t),N2(t))T and Ni(t)=exi(t) (i=1,2). Then (1.3) can be rewritten as
(2.4)x1′(t)=-a11(t)b11(t)+ex1(t)+a12(t)ex2(t)-x1(t)b12(t)+ex2(t) +∑j=1lc1j(t)ex1(t-τ1j(t))-x1(t)-γ1j(t)ex1(t-τ1j(t)):=Δ1(x,t),x2′(t)=-a22(t)b22(t)+ex2(t)+a21(t)ex1(t)-x2(t)b21(t)+ex1(t) +∑j=1lc2j(t)ex2(t-τ2j(t))-x2(t)-γ2j(t)ex2(t-τ2j(t)):=Δ2(x,t).
As usual, let X=Z={x=(x1(t),x2(t))T∈C(R,R2):x(t+ω)=x(t) for all t∈R} be Banach spaces equipped with the supremum norm ||·||. For any x∈X, because of periodicity, it is easy to see that Δ(x,·)=(Δ1(x,·),Δ2(x,·))T∈C(R,R2) is ω-periodic. Let
(2.5)L:D(L)={x∈X:x∈C1(R,R2)}∋x↦x′=(x1′,x2′)T∈Z,P:X∋x↦(1ω∫0ωx1(s)ds,1ω∫0ωx2(s)ds)T∈X,Q:Z∋z↦(1ω∫0ωz1(s)ds,1ω∫0ωz2(s)ds)T∈Z,N~:X∋x↦Δ(x,·)∈Z.
It is easy to see that
(2.6)ImL={x∣x∈Z,∫0ωx(s)ds=(0,0)T}, KerL=R2,ImP=KerL, KerQ=ImL.

Thus, the operator L is a Fredholm operator with index zero. Furthermore, denoting by LP-1:ImL→D(L)∩KerP the inverse of L|D(L)∩KerP, we have
(2.7)LP-1y(t)=-1ω∫0ω∫0ty(s)ds dt+∫0ty(s)ds=(-1ω∫0ω∫0ty1(s)ds dt+∫0ty1(s)ds,-1ω∫0ω∫0ty2(s)ds dt+∫0ty2(s)ds)T.
It follows that
(2.8)QN~x=1ω∫0ωN~x(t)dt=(1ω∫0ωΔ1(x(t),t)dt,1ω∫0ωΔ2(x(t),t)dt)T,(2.9)LP-1(I-Q)N~x=∫0tN~x(s)ds-tω∫0ωN~x(s)ds-1ω∫0ω∫0tN~x(s)ds dt +1ω∫0ω∫0tQN~x(s)ds dt.
Obviously, QN~ and LP-1(I-Q)N~ are continuous. It is not difficult to show that LP-1(I-Q)N~(Ω¯) is compact for any open bounded set Ω⊂X by using the Arzela-Ascoli theorem. Moreover, QN~(Ω¯) is clearly bounded. Thus N~ is L-compact on Ω¯ with any open bounded set Ω⊂X.

Considering the operator equation Lx=λN~x,λ∈(0,1), we have
(2.10)x′(t)=(x1′(t),x2′(t))T=λΔ(x,t)=(λΔ1(x,t),λΔ2(x,t))T.
Suppose that x=(x1(t),x2(t))T∈X is a solution of (2.10) for some λ∈(0,1).

Firstly, we claim that there exists a positive number H such that ||x||<H. Integrating the first equation of (2.10) and in view of x∈X, it results that
(2.11)0=∫0ωx1′(t)dt=λ∫0ωΔ1(x,t)dt,
which together with (2.4) implies that
(2.12)∫0ω|∑j=1lc1j(t)ex1(t-τ1j(t))-x1(t)-γ1j(t)ex1(t-τ1j(t))+a12(t)ex2(t)-x1(t)b12(t)+ex2(t)|dt =∫0ωa11(t)b11(t)+ex1(t)dt <∫0ωa11(t)b11(t)dt.
Similarly, we have
(2.13)∫0ω|∑j=1lc2j(t)ex2(t-τ2j(t))-x2(t)-γ2j(t)ex2(t-τ2j(t))+a21(t)ex1(t)-x2(t)b21(t)+ex1(t)|dt =∫0ωa22(t)b22(t)+ex2(t)dt <∫0ωa22(t)b22(t)dt.
It follows from (2.12) and (2.13) that
(2.14)∫0ω|x1′(t)|dt≤λ∫0ω|∑j=1lc1j(t)ex1(t-τ1j(t))-x1(t)-γ1j(t)ex1(t-τ1j(t))+a12(t)ex2(t)-x1(t)b12(t)+ex2(t)|dt +λ∫0ω|a11(t)b11(t)+ex1(t)|dt<2∫0ωa11(t)b11(t)dt=A1,(2.15)∫0ω|x2′(t)|dt≤λ∫0ω|∑j=1lc2j(t)ex2(t-τ2j(t))-x2(t)-γ2j(t)ex2(t-τ2j(t))+a21(t)ex1(t)-x2(t)b21(t)+ex1(t)|dt +λ∫0ω|a22(t)b22(t)+ex2(t)|dt<2∫0ωa22(t)b22(t)dt=A2.

Since x∈X, there exist ξ1,ξ2,η1,η2∈[0,ω] such that
(2.16)xi(ξi)=mint∈[0,ω]xi(t), xi(ηi)=maxt∈[0,ω]xi(t), xi′(ξi)=xi′(ηi)=0, i=1,2.
It follows from (2.12) and (2.14) that
(2.17)A12=∫0ωa11(t)b11(t)dt>∫0ωa11(t)b11(t)+ex1(t)dt=∫0ω∑j=1lc1j(t)ex1(t-τ1j(t))-x1(t)-γ1j(t)ex1(t-τ1j(t))dt+∫0ωa12(t)ex2(t)-x1(t)b12(t)+ex2(t)dt>ex1(ξ1)-x1(η1)-γ1+ex1(η1)∑j=1l∫0ωc1j(t)dt=B1ex1(ξ1)-x1(η1)-γ1+ex1(η1),
which implies that
(2.18)x1(ξ1)<lnA12B1+x1(η1)+γ1+ex1(η1).
Using (2.14) yields
(2.19)x1(t)≤x1(ξ1)+∫0ω|x1′(t)|dt<lnA12B1+x1(η1)+γ1+ex1(η1)+A1, t∈[0,ω].
In particular,
(2.20)x1(η1)<x1(ξ1)+∫0ω|x1′(t)|dt<lnA12B1+x1(η1)+γ1+ex1(η1)+A1.
It follows that
(2.21)x1(η1)>ln(1γ1+(ln2B1A1-A1)).
Again from (2.14), we have
(2.22)x1(t)≥x1(η1)-∫0ω|x1′(t)|dt>ln(1γ1+(ln2B1A1-A1))-A1:=H11, t∈[0,ω].
Similarly, we can obtain
(2.23)x2(t)≥x2(η2)-∫0ω|x2′(t)|dt>ln(1γ2+(ln2B2A2-A2))-A2:=H21, t∈[0,ω].
Since x1′(ξ1)=0, from (2.10), we have
(2.24)a11(ξ1)b11(ξ1)+ex1(ξ1)=∑j=1lc1j(ξ1)ex1(ξ1-τ1j(ξ1))-x1(ξ1)-γ1j(ξ1)ex1(ξ1-τ1j(ξ1)) +a12(ξ1)ex2(ξ1)-x1(ξ1)b12(ξ1)+ex2(ξ1).
Hence, from (2.24) and the fact that supu≥0ue-u=1/e, we have
(2.25)ex1(ξ1)b11++ex1(ξ1)≤ex1(ξ1)b11(ξ1)+ex1(ξ1)=∑j=1lc1j(ξ1)a11(ξ1)γ1j(ξ1)γ1j(ξ1)ex1(ξ1-τ1j(ξ1))e-γ1j(ξ1)ex1(ξ1-τ1j(ξ1)) +a12(ξ1)a11(ξ1)(1+b12(ξ1)e-x2(ξ1))<∑j=1lc1j+a11-γ1j-e+a12+a11-.
Noting that u/(b11++u) is strictly monotone increasing on [0,+∞) and
(2.26)supu≥0ub11++u=1>∑j=1lc1j+a11-γ1j-e+a12+a11-,it is clear that there exists a constant k1>0 such that
(2.27)ub11++u>∑j=1lc1j+a11-γ1j-e+a12+a11- ∀u∈[k1,+∞).
In view of (2.25) and (2.27), we get
(2.28)ex1(ξ1)≤k1, x1(ξ1)≤lnk1.
In the same way, there exists a constant k2>0 such that
(2.29)x2(ξ2)≤lnk2.
Again from (2.14), (2.15), (2.28), and (2.29), we get
(2.30)x1(t)≤x1(ξ1)+∫0ω|x1′(t)|dt<lnk1+A1,x2(t)≤x2(ξ1)+∫0ω|x2′(t)|dt<lnk1+A2.
Then, we can choose two sufficiently large positive constants H12>lnk1+A1 and H22>lnk2+A2 such that
(2.31)x1(t)<H12, x2(t)<H22, lnb11+<H12, lnb22+<H22.

Let H>max{|H11|,|H21|,H12,H22} be a fix constant such that
(2.32)eH>1γ1-(H-lnC1-2D12B1), eH>1γ2-(H-lnC2-2D22B2).
Then (2.22), (2.23), and (2.31) imply that ||x||<H, if x∈X is solution of (2.10). So we can define an open bounded set as Ω={x∈X:||x||<H} such that there is no λ∈(0,1) and x∈∂Ω such that Lx=λN~x. That is to say Lx≠λN~x for all x∈∂Ω∩D(L),λ∈(0,1).

Secondly, we prove that N~x∉ImL for all x∈∂Ω∩KerL. That is ((QN~(x))1,(QN~(x))2)T≠(0,0)T for all x∈∂Ω∩KerL.

If x(t)=(x1(t),x2(t))T∈∂Ω∩KerL, then x(t) is a constant vector in R2, and there exists some i∈{1,2} such that |xi|=H. Assume |x1|=H, so that x1=±H. Then, we claim
(2.33)(QN~(x))1>0 for x1=-H, (QN~(x))1<0 for x1=H.
If (QN~(x))1≤0 for x1=-H, it follows from (2.2) and (2.8) that
(2.34)∫0ωΔ1(x,t)dt≤0, for x1=-H.
Hence,
(2.35)A12=∫0ωa11(t)b11(t)dt>∫0ωa11(t)b11(t)+e-Hdt≥∫0ω[∑j=1lc1j(t)e-γ1j(t)e-H+a12(t)ex2+Hb12(t)+ex2]dt>∫0ω∑j=1lc1j(t)e-γ1j+e-Hdt≥e-γ1+e-H∑j=1l∫0ωc1j(t)dt=B1e-γ1+e-H,
which implies
(2.36)-H>ln(1γ1+ln2B1A1)>ln(1γ1+(ln2B1A1-A1))-A1=H11.
This is a contradiction and implies that (QN~(x))1>0 for x1=-H.

If (QN~(x))1≥0 for x1=H, it follows from (2.2) and (2.8) that
(2.37)∫0ωΔ1(x,t)dt≥0, for x1=H,C12e-H=∫0ωa11(t)2eHdt<∫0ωa11(t)b11(t)+eHdt≤∫0ω∑j=1lc1j(t)e-γ1j(t)eHdt+∫0ωa12(t)ex2-Hb12(t)+ex2dt≤∫0ω∑j=1lc1j(t)e-γ1-eHdt+∫0ωa12(t)b12(t)eH-x2+eHdt<e-γ1-eH∑j=1l∫0ωc1j(t)dt+∫0ωa12(t)eHdt=B1e-γ1-eH+D1e-H.
Consequently,
(2.38)eH<1γ1-(H-lnC1-2D12B1),
a contradiction to the choice of H. Thus, (QN~(x))1<0 for x1=H.

Similarly, if |x2|=H, we obtain
(2.39)(QN~(x))2>0 for x2=-H, (QN~(x))2<0 for x2=H.
Consequently, (2.33) and (2.39) imply that N~x∉ImL for all x∈∂Ω∩KerL.

Furthermore, let 0≤μ≤1 and define continuous function H(x,μ) by setting
(2.40)H(x,μ)=-(1-μ)x+μQN~x.

For all x(t)=(x1(t),x2(t))T∈∂Ω∩KerL, then there exists some i∈{1,2} such that |xi|=H. There are two cases: x1=±H or x2=±H. When x1=H or x2=H, from (2.33) and (2.39), it is obvious that (H(x,μ))1<0 or (H(x,μ))2<0. Similarly, if x1=-H or x2=-H, it results that (H(x,μ))1>0 or (H(x,μ))2>0. Hence H(x,μ)≠(0,0)T for all x∈∂Ω∩kerL.

Finally, using the homotopy invariance theorem, we obtain
(2.41)deg{QN~,Ω∩kerL,(0,0)T}= deg{-x,Ω∩kerL,(0,0)T}≠0.
It then follows from the continuation theorem that Lx=N~x has a solution
(2.42)x*(t)=(x1*(t),x2*(t))T∈DomL⋂Ω¯,
which is an ω-periodic solution to (2.4). Therefore N*(t)=(N1*(t),N2*(t))T=(ex1*(t),ex2*(t))T is a positive ω-periodic solution of (1.3) and the proof is complete.