Yay for 2014! 5

We can write: $14k \equiv 14 \pmod{20}$ where k is an integer. Which simplifies to $k \equiv 1 \pmod{10}.$

Now, from the second condition, we write: $14k \equiv 4 \pmod{201}$ $k \equiv 29 \pmod{201}.$

The smallest k that satisfies these conditions is k=431. Then, our number is 431*14=6034.

Therefore, the sum of the digits is $6+0+3+4=\boxed{13}$ .

$From\ the\ given\ data\ we\ derive\ the\ following\ equations:\\ 14x-14=20a\\ 14x-4=201b.\\Now,subtract\ the\ above\ two\ equations\ to\ get:\\ 10=20a-201b\\ \Rightarrow10(1+2a)=201b.\\Now,because\ gcd(201,10)=1,\\the\ values\ of\ (1+2a)\ are\ of\ the\ form\ 201y.Thus,(1+2a)=201y.\\Substituting\ this\ value\ in\ the\ first\ equation\ gives:\\ 14x-14=2010y-10\\ \Rightarrow 14x=2010y+4.\\ But,2010=14p+8(p=143)\\ \Rightarrow 14x=14ay+8y+4.\\ \Rightarrow 8y+4\ must\ be\ divisible\ by\ 14.The\ smallest\ value\ of\ y\ that\ satisfies\ this\ \\is\ 3.\\ \Rightarrow 14x=6034.$

Here's a simple Python 3.0 script to achieve the same:

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n = 10
digit_sum = 0
while(True):
    #this condition is because on dividing by 20 it leaves remainder
    #as 14, so its last digit should be 4. Also on dividing by 201, it
    #leaves remainder as 4 so the previous multiple of 201 must be obtained by
    #multiplying 201 with a multiple of 10. However, we increase by 20 every
    #because on increase of 10 we would obtain a number which on dividing
    #by 20 leaves 4 as remainder. So, we consider only odd multiples of 10.
    #Now, two conditions are covered adding 4 and checking mod against 14 
    #checks for 3rd condition.
    if((n*201+4)%14==0):
        break
    n = n + 20
num = n*201+4
while(num>0):
    digit_sum = digit_sum + num%10
    num = num//10
print(digit_sum)