The sum of the digits of a four-digit number does not exceed 36, so for a delightful number, it must be 25. Since a delightful number is divisible by 25, it ends in either 00, 50, 25, or 75. If a four-digit number ends in 00 or 50, the sum of its digits does not exceed 9 + 9 + 5 = 23, which is not suitable. If a delightful number ends in 25, the sum of the first two digits must be 18. This could only be the number 9925, but its product of digits is not divisible by 25. Therefore, a delightful number can only end in 75. In this case, the sum of its first two digits is 13, and their product must be divisible by 5. Thus, these two digits are 5 and 8 (in any order).

Transform the given sum: \( 13^{2013} + 13^{2014} + 13^{2015} = 13^{2013}(1 + 13 + 13^2) \). Notice that \( 13^2 = 169 \equiv 46 \pmod{61} \), so \( 1 + 13 + 13^2 \equiv 1 + 13 + 46 \equiv 60 \equiv -1 \pmod{61} \). Therefore, \( 13^{2013}(1 + 13 + 13^2) \equiv 13^{2013} imes (-1) \equiv -13^{2013} \equiv 0 \pmod{61} \), since \( 13^{2013} \equiv 0 \pmod{61} \). Thus, the sum is divisible by 61.

Assume each \(2 \times 2\) square contains \(m\) pieces, and each \(1 \times 3\) rectangle contains \(n\) pieces. Consider a \(2 \times 6\) rectangle on the board. On one hand, this rectangle can be divided into three \(2 \times 2\) squares, so it contains \(3m\) pieces. On the other hand, it can be divided into four \(1 \times 3\) rectangles, so it contains \(4n\) pieces. We get the equation \(3m = 4n\), implying \(n\) is divisible by 3. Since \(n\) can be 0, 1, 2, or 3, we have \(n = 0\) or \(n = 3\). In other words, either all \(1 \times 3\) rectangles are empty, resulting in 0 pieces on the board, or all \(1 \times 3\) rectangles are fully occupied, resulting in 64 pieces on the board.

Method 1 (finding numbers). Let the sum of numbers from a to a + 2014 end with the same digit as the sum of numbers from a + 2015 to a + 4033. Then the difference between these sums, equal to (a + 2015 + a + 4033) - (a + a + 2014), must be divisible by 10. This simplifies to 2015 + 4033 - 2014, which equals 2019. Since 2019 is odd, it cannot be divisible by 10. Contradiction.

Method 2 (parity directly). Among 2015 + 2019 = 4034 consecutive natural numbers, there are 2017 odd numbers, which is an odd count. One sum contains an even number of odd numbers, and the other contains an odd number. Therefore, the sums will have different parity, meaning they cannot end with the same digit.

Answer. No, it cannot.

Let \( x \) be the left number and \( y \) be the right number; according to the condition: \( x + y = S \).

Then Masha obtained the number \( ax + by \), and Alyosha obtained the number \( bx + ay \).

The sum of these numbers is \( ax + by + bx + ay = a(x + y) + b(x + y) = (a + b)S \), which is divisible by \( S \).

Since one of the two terms (Masha's number) is divisible by \( S \), the other (Alyosha's number) must also be divisible by \( S \), as required.

Note that the number 288 has divisors 1, 2, 3, 4, 32, 6, 16, 8, 9. Therefore, this number satisfies the problem's condition. We will prove that there is no smaller number satisfying the condition. Indeed, since \( N \) must have a divisor with a digit sum of 9, \( N \) is divisible by 9. Now consider a divisor \( d \) with a digit sum of 8. \( d \) is not divisible by 3, so \( d \) and 9 are coprime, meaning \( N \) is divisible by \( 9d \). If \( d \geq 32 \), then \( 9d \geq 288 \), which implies \( N \geq 288 \). Thus, we need to check \( d = 26 \), \( d = 17 \), and \( d = 8 \). If \( d = 26 \), then \( 9d = 234 \). This number does not have a divisor with a digit sum of 5, and any number that is a multiple of it is greater than 288. If \( d = 17 \), then \( 9d = 153 \). This number does not have a divisor with a digit sum of 2, and any number that is a multiple of it is greater than 288. If \( d = 8 \), then \( 9d = 72 \). Its multiples less than 288 are 144 and 216. But these numbers do not have a divisor with a digit sum of 5.