Let $M$ be the set of palindromic integers of the form $5n+4$ where $n\ge 0$ is an integer. 
[list=a]
[*]If we write the elements of $M$ in increasing order, what is the $50^{\text{th}}$ number?
[*]Among all numbers in $M$ with nonzero digits which sum up to $2014$ which is the largest and smallest one?[/list]
### Part (a)
To find the $50^{\text{th}}$ number in the set $M$ of palindromic integers of the form $5n + 4$, we need to generate the first 50 palindromic numbers that satisfy this form.

1. **Identify the form of palindromic numbers**: A palindromic number reads the same forwards and backwards. Examples include 4, 9, 11, 22, 33, etc.
2. **Check the form $5n + 4$**: We need to find palindromic numbers that can be expressed as $5n + 4$ for some integer $n \ge 0$.
3. **Generate palindromic numbers and check**: Start generating palindromic numbers and check if they can be written in the form $5n + 4$.

Let's list the first few palindromic numbers and check:
- 4: $4 = 5(0) + 4$
- 9: $9 = 5(1) + 4$
- 22: $22 = 5(4) + 4$
- 33: $33 = 5(5) + 4$
- 44: $44 = 5(8) + 4$
- 55: $55 = 5(11) + 4$
- 66: $66 = 5(12) + 4$
- 77: $77 = 5(15) + 4$
- 88: $88 = 5(16) + 4$
- 99: $99 = 5(19) + 4$

Continue this process until we find the 50th palindromic number that fits the form $5n + 4$.

### Part (b)
To find the largest and smallest numbers in $M$ with nonzero digits summing to 2014:

1. **Largest number**:
   - We want as many digits as possible, so we start and end with 4.
   - The sum of the digits is 2014, so the middle digits must sum to 2014 - 4 - 4 = 2006.
   - Use 2006 ones in the middle to maximize the number of digits.

   Thus, the largest number is:
   \[
   \boxed{4\underbrace{111\ldots111}_{2006 \text{ ones}}4}
   \]

2. **Smallest number**:
   - We want as few digits as possible, so we start and end with 9.
   - The sum of the digits is 2014, so the middle digits must sum to 2014 - 9 - 9 = 1996.
   - Use as many 9s as possible in the middle. The maximum number of 9s is $\left\lfloor \frac{1996}{9} \right\rfloor = 222$.
   - The remaining sum is 1996 - 222 \times 9 = 1996 - 1998 = -2, so we need to adjust by using 7s.

   Thus, the smallest number is:
   \[
   \boxed{9\underbrace{999\ldots999}_{110 \text{ nines}}77\underbrace{999\ldots999}_{110 \text{ nines}}9}
   \]