The $a$ in the sum above should be an $n$. Lemma 1: There exists a $C$ such that $a_n < C/n$ for all positive integers $n$. Proof: Since $\{ a_n \}$ is bounded, say $a_n < D$ for all $n$. Then \[a_n < \frac{1}{2n+2007} + \sum_{k=n}^{2n+2006} \frac{a_k}{k+1} < \frac{1}{2n+2007} + D \cdot \ln \frac{2n+2007}{n} < \frac{1}{2n+2007} + D \cdot \ln 2.5\] (using that $1+\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \approx \ln n.$ ) for any $n > 9001$. If we continue iterating the above operation for all $a_n$ for many many steps, we can show that it approaches some limit. (since it is decreasing, but always stays positive). What limit does it approach? Essentially I am applying the operation $a_n \rightarrow \frac{1}{2n+2007} + a_n \cdot \ln 2.5$ $(*)$, (this works mostly because $\frac{1}{2n+2007} + D \cdot \ln 2.5 > \frac{1}{2(n+1)+2007} + D \cdot \ln 2.5$ ) If we continue iterating as in $(*)$ you eventually reach $O(n^{-1}).$ Now say that $a_n < C/n$ for all $n$ for some $C$. I'll show that we can actually get to $C = 1.$ Assume that $C = 1 + \epsilon$ and that $a_n = C/n$ for some $n$. Then \[a_n < \frac{1}{2n+2007} + \sum_{k=n}^{2n+2006} \frac{a_k}{k+1} < \frac{1}{2n+2007} + C \left( \frac{1}{n} - \frac{1}{2n+2007} \right) \] by telescoping. Similar to above, we iterate this inequality for all values of $n.$ In particular, if $a_n < C/n$ for all $n$, then $a_n < \left( \frac{n}{2n+2007} + Cn \left( \frac{1}{n} - \frac{1}{2n+2007} \right) \right)/n$ for any $n$. The quantity on the right is decreasing in $n$ for $C > 1$. So this allows us to iterate $C \rightarrow \frac{n}{2n+2007} + Cn \left( \frac{1}{n} - \frac{1}{2n+2007} \right).$ One can check that the limit of $C$ is $1$. So if $C > 1 + \epsilon$, for enough iterations, this will be false. So $a_n \le 1/n$ for all $n$. Using the condition one more time and telescoping finishes.

I guess it would be natural for many people to make the substitute $b_n:=a_n-\frac{1}{n}$ as a first step, then we have \[b_n<\sum_{k=n}^{2n+2006}[\frac{b_k}{k+1}-\frac{1}{k(k+1)}]+\frac{1}{2n+2007}-\frac{1}{n}=\sum_{k=n}^{2n+2006}\frac{b_k}{k+1},\forall n\geq 1.\]Define $M_n:=\sup_{k\geq n}b_k$ for $n\geq 1$, then by assumption $M_n<+\infty$ and for any $m\geq n$, $b_m<\sum_{k=m}^{2m+2006}\frac{M_n}{k+1}$. If $M_n>0$, then as \[\sum_{k=n}^{2n+2006}\frac{1}{k+1}<\int_n^{2n+2007}\frac{dx}{x}=\ln(2+\frac{2007}{n})\]it holds that $M_n\leq\ln(2+\frac{2007}{n})M_n$. Thus when $n>\frac{2007}{e-2}$ we have $M_n\leq 0$ and $b_n<0$. On the other hand we have \[\frac{n}{n+1}b_n<\sum_{k=n+1}^{2n+2006}\frac{b_k}{k+1},\forall n\geq 1,\]so by backward induction we have $b_n<0$ for all $n\geq 1$.

First set $b_n:=a_n-\dfrac1n$.Then we can easily get $b_n<\sum_{k=n}^{2n+2006}\dfrac{b_k}{k+1}$. Suppose $b_n\ge0$,then $\sum_{k=n+1}^{2n+2006}\dfrac{b_k}{k+1}>\dfrac n{n+1}b_n\ge0$ so $\exists k \in [n+1,2n+2006],b_k>0$. Repeat this then we can know there exists a $m\ge 100000$ that $b_m>0$. Set $\epsilon:= b_m$ and $\alpha:=\dfrac1{\ln{2.6}}$. Here $\alpha >1$. For $n\ge 4$ we have $1+\dfrac12+\dfrac13+\dfrac14+\int_5^{n+1}\dfrac{\text{d}x}x\le\sum_{i=1}^n\dfrac1i\le1+\dfrac12+\dfrac13+\dfrac14+\int_4^n\dfrac{\text{d}x}x$ so $\sum_{k=m}^{2m+2006}\dfrac1{k+1}=\sum_{k=1}^{2m+2007}\dfrac1k-\sum_{k=1}^m\dfrac1k\le \left(1+\dfrac12+\dfrac13+\dfrac14+\int_4^{2m+2007}\dfrac{\text{d}x}x\right)-\left(1+\dfrac12+\dfrac13+\dfrac14+\int_5^{m+1}\dfrac{\text{d}x}x\right)=\ln\left(\dfrac{10m+10035}{4m+4}\right)=\ln\left(2.6-\dfrac{0.4m-10024.6}{4m+4}\right)<\ln 2.6=\dfrac1\alpha$. If $\forall k\in[n+1,2n+2006],b_k\le \alpha \epsilon$ then $\epsilon=b_n<\sum_{k=n}^{2n+2006}\dfrac{b_k}{k+1}< \dfrac1\alpha \times \alpha\epsilon=\epsilon$,a contradiction. So $\exists k\in[n+1,2n+2006],b_k>\alpha\epsilon$. Repeat this then we can find $b_m>\alpha^n\epsilon$ for every positive integer $n$ so $\{b_n\}$ is not bounded and so does $\{a_n\}$,a contradiction. So $b_n<0$ for every positive integer $n$,so $a_n<\dfrac 1n$.

The analogous problem where $<$ is replaced by $=$ appears on Bulgaria IMO TST 2017, with the aim to show that $a_n = \frac{1}{n}$.