2. How many pairs of positive integers $(x, y)$ are there such that $x<y$, and $\frac{x^{2}+y^{2}}{x+y}$ is a divisor of 2835?
2. Notice that, $2835=3^{4} \times 5 \times 7$.

Let $\frac{x^{2}+y^{2}}{x+y}=k$ (where $k$ is a factor of 2835). Then $x^{2}+y^{2}=k(x+y)$.
When $k=3$, if $x$ and $y$ are not divisible by 3, then their sum of squares cannot be divisible by 3, which is a contradiction; if one of $x$ and $y$ is divisible by 3, then the other must also be divisible by 3, thus, $x$ and $y$ are both divisible by 3.
Similarly, when $k=7$, $x$ and $y$ are both divisible by 7.
Therefore, we can eliminate all factors of 3 and 7, resulting in $x^{2}+y^{2}=k^{\prime}(x+y)\left(k^{\prime}=1\right.$ or 5$)$.
If $k^{\prime}=1$, then
$$
(2 x-1)^{2}+(2 y-1)^{2}=2 \text {, }
$$

there is no solution that satisfies $x<y$.
If $k^{\prime}=5$, then
$$
(2 x-5)^{2}+(2 y-5)^{2}=50
$$
$\Rightarrow(x, y)=(3,6)$ or $(2,6)$.
Thus, the solutions to $x^{2}+y^{2}=k(x+y)$ that satisfy $x<y$ are $(x, y)=\left(\frac{3}{5} k, \frac{6}{5} k\right)$ or $\left(\frac{2}{5} k, \frac{6}{5} k\right)(5 \mid k)$.
Since 2835 has 10 positive factors divisible by 5, therefore, there are 20 pairs of positive integers $(x, y)$ in total.