Problem 1579

1. Case 1: 2-digit numbers

We need to find all 2-digit numbers whose digits sum to 10. Listing them:
$$19, 28, 37, 46, 55, 64, 73, 82, 91$$
This gives us a total of 9 numbers.

2. Case 2: 3-digit numbers

We need to find all 3-digit numbers whose digits sum to 10. We can break this down by the hundreds digit:
- If the hundreds digit is 1, the remaining two digits must sum to 9. The valid combinations are:
$$109, 118, 127, 136, 145, 154, 163, 172, 181, 190$$
This gives us 10 numbers.
- If the hundreds digit is 2, the remaining two digits must sum to 8. The valid combinations are:
$$208, 217, 226, 235, 244, 253, 262, 271, 280$$
This gives us 9 numbers.
- If the hundreds digit is 3, the remaining two digits must sum to 7. The valid combinations are:
$$307, 316, 325, 334, 343, 352, 361, 370$$
This gives us 8 numbers.
- If the hundreds digit is 4, the remaining two digits must sum to 6. The valid combinations are:
$$406, 415, 424, 433, 442, 451, 460$$
This gives us 7 numbers.
- If the hundreds digit is 5, the remaining two digits must sum to 5. The valid combinations are:
$$505, 514, 523, 532, 541, 550$$
This gives us 6 numbers.
- If the hundreds digit is 6, the remaining two digits must sum to 4. The valid combinations are:
$$604, 613, 622, 631, 640$$
This gives us 5 numbers.
- If the hundreds digit is 7, the remaining two digits must sum to 3. The valid combinations are:
$$703, 712, 721, 730$$
This gives us 4 numbers.
- If the hundreds digit is 8, the remaining two digits must sum to 2. The valid combinations are:
$$802, 811, 820$$
This gives us 3 numbers.
- If the hundreds digit is 9, the remaining two digits must sum to 1. The valid combinations are:
$$901, 910$$
This gives us 2 numbers.

Summing these, we get:
$$10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 = 54$$

3. Case 3: 4-digit numbers

We need to find all 4-digit numbers whose digits sum to 10. We can break this down by the thousands digit:
- If the thousands digit is 1, the remaining three digits must sum to 9. The valid combinations are:
$$1009, 1018, 1027, 1036, 1045, 1054, 1063, 1072, 1081, 1090, 1108, 1117, 1126, 1135, 1144, 1153, 1162, 1171, 1180, 1207, 1216, 1225, 1234, 1243, 1252, 1261, 1270, 1306, 1315, 1324, 1333, 1342, 1351, 1360, 1405, 1414, 1423, 1432, 1441, 1450, 1504, 1513, 1522, 1531, 1540, 1603, 1612, 1621, 1630, 1702, 1711, 1720, 1801, 1810, 1900$$
This gives us 55 numbers.
- If the thousands digit is 2, the remaining three digits must sum to 8. The valid combinations are:
$$2008, 2017, 2026, 2035, 2044, 2053, 2062, 2071, 2080$$
This gives us 9 numbers.

Summing these, we get:
$$55 + 2 = 57$$

4. Total number of valid numbers

Summing the valid numbers from all cases, we get:
$$9 + 54 + 57 = 120$$

5. Probability calculation

The total number of positive integers less than or equal to 2020 is 2020. Therefore, the probability that the digits in Wendy's number add up to 10 is:
$$\frac{120}{2020} = \frac{6}{101}$$

6. **Finding $m + n$**

Since the probability is $\frac{6}{101}$, we have $m = 6$ and $n = 101$. Therefore:
$$m + n = 6 + 101 = 107$$

The final answer is $\boxed{107}$