4. Divide a regular 2017-gon into 2015 triangular regions using 2014 non-intersecting diagonals inside it. Find the maximum possible number of isosceles triangles among these 2015 triangles.
(Hu Xiaojun, problem contributor)
4. The maximum possible value is 2010.

Let this 2017-gon be $A_{1} A_{2} \cdots A_{2017}$.
First, connect the diagonals $A_{1} A_{3}, A_{3} A_{5}, \cdots, A_{2015} A_{2017}$, to get 1008 isosceles triangles;

Then, connect the diagonals $A_{1} A_{5}, A_{5} A_{9}, \cdots, A_{2013} A_{2017}$, to get 504 isosceles triangles.

Continuing in this manner, on the $k$-th step, connect the diagonals of the form $A_{2^{k}(n-1)+1} A_{2^{k} n+1}$, and finally, we get
\[
\begin{array}{l}
1008+504+252+126+63+31+15+ \\
7+3+1=2010
\end{array}
\]
isosceles triangles.
Next, we prove: the number of isosceles triangles is no more than 2010, i.e., at least five triangles are not isosceles.

Consider a diagonal of a regular 2017-gon and the part on one side of it. If it exactly contains $k$ $(k \leqslant 1008)$ sides of the original 2017-gon, then it is called a “$k$-bow”. 

In particular, when $k=1$, there is a similar definition, at which point it degenerates into a line segment.

Let the positive integer $n$ have $s(n)$ ones in its binary representation.
First, we prove a lemma.
Lemma: After dividing a $k$-bow into triangles using diagonals, there are at least $s(k)-1$ triangles that are not isosceles.
Proof: When $k=1,2$, the conclusion is obviously true.
If the conclusion holds for all $k < t$, consider $k=t$.
Analyze the triangle containing the longest side of the $k$-bow.
(i) If it is an isosceles triangle.
Since this longest side can only be the base of an isosceles triangle, $k$ must be even, and the $k$-bow is divided into one isosceles triangle and two $\frac{k}{2}$-bows.

By the induction hypothesis, we know that there are at least $2\left(s\left(\frac{k}{2}\right)-1\right)$ triangles that are not isosceles.
Since $s\left(\frac{k}{2}\right)=s(k)$, we have
$2 s\left(\frac{k}{2}\right)-2 \geqslant s(k)-1$.
Thus, the conclusion holds.
(ii) If it is not an isosceles triangle.
Assume that the $k$-bow is divided into one triangle, one $p$-bow, and one $q$-bow.
Thus, $p+q=k$.
Since $s(p)+s(q)-s(p+q)$ represents the number of carries in the binary addition of $p+q$, we have
$s(p)+s(q) \geqslant s(k)$.
By the induction hypothesis, we know that there are at least
\[
1+(s(p)-1)+(s(q)-1) \geqslant s(k)-1
\]
triangles that are not isosceles.
Thus, the conclusion holds.
The lemma is proved.
Finally, consider the original 2017-gon.
Since no diagonal passes through the center, we can assume that the triangle containing the center is $\triangle A B C$.
Note that, $2017=(11111100001)_{2}$.
Thus, $s(2017)=7$.
(1) If $\triangle A B C$ is an isosceles triangle, let the remaining part be two $p$-bows and one $q$-bow.
Thus, $2 p+q=2017$.
Note that,
\[
\begin{array}{l}
s(2 p)=s(p), \\
s(2017) \leqslant s(2 p)+s(q) . \\
\text { By the lemma, we know that there are at least } \\
2(s(p)-1)+(s(q)-1) \\
=(s(2 p)+s(q))+(s(p)-1)-2 \\
\geqslant s(2017)-2=5
\end{array}
\]
triangles that are not isosceles.
(2) If $\triangle A B C$ is not an isosceles triangle, let the remaining part be one $p$-bow, one $q$-bow, and one $r$-bow.
Thus, $p+q+r=2017$.
Note that, $s(2017) \leqslant s(p)+s(q)+s(r)$.
By the lemma, we know that there are at least
\[
\begin{array}{l}
1+(s(p)-1)+(s(q)-1)+(s(r)-1) \\
\geqslant s(2017)-2=5
\end{array}
\]
triangles that are not isosceles.
In summary, the problem is proved.