Answer: (3020,6043$), (4027,4029), (6041,3022)$.

Solution. Transform the equation to the form $(x-2013)(y-2015)=2013 \cdot 2015+1$, the right side can be factored into primes as $2014^{2}=2^{2} \cdot 19^{2} \cdot 53^{2}$.

The right side can be decomposed into the product of two natural factors in 27 different ways, but only in seven of these ways will both factors be four-digit numbers (rows in the table without dashes):

| $x-2013$ | $y-2015$ | $x$ | $y$ | $x-2013$ | $y-2015$ | $x$ | $y$ | $x-2013$ | $y-2015$ | $x$ | $\bar{y}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | $2^{2} \cdot 19^{2} \cdot 53^{2}$ | - | more than 4-digit | $19^{2}$ | $2^{2} \cdot 53^{2}$ | 2374 | - | 53 | $2^{2} \cdot 19^{2} \cdot 53$ | - | - |
| 2 | $2 \cdot 19^{2} \cdot 53^{2}$ | - |  | $2 \cdot 19^{2}$ | $2 \cdot 53^{2}$ | 2735 | 7633 | $2 \cdot 53$ | $2 \cdot 19^{2} \cdot 53$ | - | - |
| $2^{2}$ | $19^{2} \cdot 53^{2}$ | - |  | $2^{2} \cdot 19^{2}$ | $53^{2}$ | 3457 | 4824 | $2^{2} \cdot 53$ | $19^{2} \cdot 53$ | - | - |
| 19 | $2^{2} \cdot 19 \cdot 53^{2}$ | - |  | $53^{2}$ | $2^{2} \cdot 19^{2}$ | 4822 | 3459 | $19 \cdot 53$ | $2^{2} \cdot 19 \cdot 53$ | 3020 | 6043 |
| $2 \cdot 19$ | $2 \cdot 19 \cdot 53^{2}$ | - |  | $2 \cdot 53^{2}$ | $2 \cdot 19^{2}$ | 7631 | 2737 | $2 \cdot 19 \cdot 53$ | $2 \cdot 19 \cdot 53$ | 4027 | 4029 |
| $2^{2} \cdot 19$ | $19 \cdot 53^{2}$ | - |  | $2^{2} \cdot 53^{2}$ | $19^{2}$ | - | 2376 | $2^{2} \cdot 19 \cdot 53$ | $19 \cdot 53$ | 6041 | 3022 |

Criteria. Losing part of the solutions: minus 3 points.
10.3. Find all solutions in four-digit natural numbers for the equation $1+2013 x+2015 y=x y$.