Main idea : Try each digit sum from the smallest, and see $53$ doesn't divide numbers with digit sum less than $3$ . Also brute force $10^n \bmod 53$ by computer.

Complete proof :

Note that $2014 = 2 \cdot 19 \cdot 53$ .

First, note that $10^{23} + 10^{22} + 10$ is divisible by $2014$ (if you like working, you can compute by hand; I personally used Python), and it has digit sum $3$ . So we get an upper bound of the minimum to be $3$ . Now we will prove that a digit sum of $1$ or $2$ is impossible.

If the digit sum is $1$ , then our number must be in the form of $10^k = 2^k5^k$ , not containing the factor $19$ or $53$ . So this is clearly impossible.

If the digit sum is $2$ , then our number must be in the form of $10^a + 10^b$ ; WLOG assume that $a \ge b$ . So our number is in the form $10^b (10^k + 1)$ where $k = a-b \ge 0$ . Again, since $10^b = 2^b5^b$ doesn't contain the factor $53$ , then $10^k+1$ must contain the factor $53$ ; in other words, $10^k+1 \equiv 0 \pmod{53}$ , or $10^k \equiv -1 \equiv 52 \pmod{53}$ .

However, we observe that $10^{13} \equiv 1 \pmod{53}$ (Python), so the residues upon dividing by $53$ will be periodic with period $13$ . Also, we can verify that $10^k \equiv 1, 10, 47, 46, 36, 42, 49, 13, 24, 28, 15, 44, 16 \pmod{53}$ (Python). Since we have $13$ residues, these are all possible residues, and it can be trivially seen that $52$ is not among those residues, so there is no integer $k$ making $10^k \equiv -1 \equiv 52 \pmod{53}$ . In other words, $10^k+1$ is never divisible by $53$ .

So our number $10^a+10^b$ is not divisible by $53$ , and hence the digit sum $2$ fails. So the lower bound of the minimum is $3$ . As we have shown a model that $3$ works, our answer is thus $\boxed{3}$ .

Motivation (or my solving path) :

Since we want to find the smallest digit sum, I went up from the smallest possible digit sum. The smallest, $1$ , was trivially discarded. After that, initially I thought $10^k+1$ takes all residues for any integer $N$ not divisible by $2$ or $5$ , so I submitted $2$ , and got wrong. So then I factorized $2014 = 2 \cdot 19 \cdot 53$ , and started coding, to figure out that $10^k \bmod 53$ repeats with smallest period $13$ and not $\varphi(53) = 52$ as expected. The rest of this path is described in the proof above pretty identically. So that's out of the window too.

Then I went up to $3$ , which gives so much freedom in the form of $10^a + 10^b + 1$ , not to mention that the period of $10^k \pmod{19}$ is indeed $\varphi(19) = 18$ , spewing all possible residues except $0$ . So if we can figure out $a,b$ satisfying $10^a + 10^b + 1 = 0 \pmod{53}$ , I can simply increase $a$ by a multiple of $13$ , keeping $10^a + 10^b + 1 = 0 \pmod{53}$ , but at the same time rolling over all residues of $19$ , pretty much guaranteeing a hit. Just for fun, I tried to find the smallest number satisfying the condition; I believe $10^{23} + 10^{22} + 10$ is indeed the smallest among all having digit sum $3$ .

Generalization :

The obvious generalization is changing $2014$ . We can discard any factor of $2$ or $5$ in the divisor by appending enough $0$ 's at the end, so we are left with several prime factors, none of them is $2$ or $5$ . After this, I think there's no way to proceed except by first brute forcing $10^n \bmod d$ for each prime divisor $d$ of $N$ , then trying to figure out some $a,b$ (or more) that makes the sum correct, hoping that you don't hit a bad divisor like $37$ (whose residues are $1,10,26$ only!) that makes your problem more insane...

Let's start with numbers with the smallest digit sum and see where it leads us. For digit sum $1$ the only possibility is $N = 10^k$ for some $k$ , which is impossible because $2014$ has prime factors different from $2$ and $5$ .

For digit sum $2$ there are two possibilities. Either $N = 2 \cdot 10^k$ which doesn't work for the same reason as above or $N= 10^k + 10^l$ . In this case, observe that there is no factor of $5$ in $2014$ , so that we can divide $N$ by $10^l$ and study instead the divisibility of $N' = 10^n + 1$ by $2014/2 = 1007$ . Because $1007 = 19 \cdot 53$ , by chinese remainder theorem we have to solve the congruences $10^n + 1 \equiv 0 \pmod{19},$ $10^n + 1 \equiv 0 \pmod{53}.$ The only remainders of $10^n$ modulo $53$ are $1, 10, 47, 46, 36, 42, 49, 13, 24, 28, 15, 44, 16$ , so there is no solution for digit sum equal to $2$ .

We failed previously but we learned something valuable. Namely, that it's enough to investigate simple congruences modulo $19$ and $53$ . We notice above that $42$ is a remainder of $10^n$ , so that $10^n + 11 \equiv 0 \pmod{53}$ does have a solution. And there is a solution modulo $19$ as well, as $10^n$ generates all of $18$ non-zero elements of ${\bf Z} / 19 {\bf Z}$ .

Putting this together, there is $n$ such that $10^n + 11$ is divisible by both $19$ and $53$ and therefore also by $1007$ . Therefore $10^{n+1} + 110$ is divisible by $2014$ and the answer is $\bf 3$ .

Souryajit Roy 2014=2x19x53. Consider the powers of 10 mod 53.They are as follows 10, 47, 46, 36, 42, 49, 13, 24, 28, 15, 44, 16, 1. Consider the powers of 10 mod 19.They are : 10, 5, 12, 6, 3, 11, 15, 17, 18, 9, 14, 7, 13, 16, 8, 4, 2, 1 . The powers of 10 mod 2 are 1 or 0 Since neither 0 nor 52 occurs in the list for 53,the minimum digit sum is 3. 10^52=1(mod 53) by Fermat's theorem and hence 10^13n=1(mod 53) if n is a multiple of 4. Also see 10^5=42(mod 53).Therefore, 10^(13n+5)=42(mod 53) if n is a multiple of 4. Hence, 1+10+10^(13n+5)=53=0(mod 53). The above expression is also divisible by 19 if 10^(13n+5)=8(mod 19). By Fermat's theorem 10^18=1(mod 19) and hence 10^18n=1(mod 19). Note that 10^5=3(mod 19) and so 10^15=3^3=8(mod 19). Hence,10^k=8(mod 19) if k=15(mod 18). Hence,we require 13n+5=15(mod 18) with a solution n=16,a multiple of 4. It follows that 1 + 10 + 10^(5+13*16) = 1 + 10 + 10^213 is divisible by both 19 and 53. At last,10 + 10^2 + 10^214 is divisible by 2, 19, and 53, and therefore it is divisible by 2014. The digit sum for this number is 3.

I found a solution with 5. Then I tried 4 and finally 3 !!!