The real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ are the consecutive terms of an arithmetic sequence. If

$$
\frac{x_{2}}{x_{1}+x_{3}}+\frac{x_{3}}{x_{2}+x_{4}}+\frac{x_{4}}{x_{3}+x_{5}}+\cdots+\frac{x_{n-2}}{x_{n-3}+x_{n-1}}+\frac{x_{n-1}}{x_{n-2}+x_{n}}=1957
$$

what is the value of $n$ ?
Since the numbers $x_{1}, x_{2}, \ldots, x_{n}$ form an arithmetic sequence, then for each integer $k$ with $2 \leq k \leq n-1$, we have $x_{k}-x_{k-1}=x_{k+1}-x_{k}$.

Rearranging, we obtain $2 x_{k}=x_{k-1}+x_{k+1}$ and so $\frac{x_{k}}{x_{k-1}+x_{k+1}}=\frac{1}{2}$ for each integer $k$ with $2 \leq k \leq n-1$.

We note that there are $(n-1)-2+1=n-2$ integers $k$ in this range.

Therefore, starting with the given equation

$$
\frac{x_{2}}{x_{1}+x_{3}}+\frac{x_{3}}{x_{2}+x_{4}}+\cdots+\frac{x_{n-2}}{x_{n-3}+x_{n-1}}+\frac{x_{n-1}}{x_{n-2}+x_{n}}=1957
$$

we obtain $(n-2) \cdot \frac{1}{2}=1957$ which gives $n-2=3914$ and so $n=3916$.

ANSWER: 3916