3. Let $m$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{0}=m, a_{1}=\varphi(m), \\
a_{2}=\varphi^{(2)}(m)=\varphi(\varphi(m)), \cdots, \\
a_{n}=\varphi^{(n)}(m)=\varphi\left(\varphi^{(n-1)}(m)\right),
\end{array}
$$

where $\varphi(m)$ is the Euler's function.
If for any non-negative integer $k$, we have $a_{k+1} \mid a_{k}$, find the largest positive integer $m$ not exceeding 2016.
3. It is known that if $a \mid b$, then $\varphi(a) \mid \varphi(b)$.

Moreover, when $m>2$, $\varphi(m)$ is even.
Therefore, it is only necessary to find integers $m$ such that $a_{1} \mid a_{0}$, which implies $a_{k+1} \mid a_{k}$.
If $m$ is an odd number greater than 2, it is impossible for $a_{1} \mid m$.
Let $m=2^{\alpha} \prod_{k=1}^{s} p_{k}^{\alpha_{s}}$, where $p_{1}, p_{2}, \cdots, p_{s}$ are all odd primes. Then
$$
\varphi(m)=m\left(1-\frac{1}{2}\right) \prod_{k=1}^{s}\left(1-\frac{1}{p_{k}}\right) .
$$

Thus, only when $s=1$ and $p_{1}=3$, we have $a_{1} \mid a_{0}$.
Therefore, $m$ must be of the form $2^{\alpha} \times 3^{\beta}$.
Hence, the largest positive integer value not exceeding 2016 is
$$
m=2^{3} \times 3^{5}=1944 \text {. }
$$