Minimum value 1

Let $n={ 2 }^{ k }\times t$ , $k,t$ are whole numbers, and $t$ is odd.

We have ${ 2013 }^{ n }-1={ { \left( { 2013 }^{ 2^{ k } } \right) }^{ t }-1=\left( { 2013 }^{ 2^{ k } }-1 \right) }\left[ { \left( { 2013 }^{ 2^{ k } } \right) }^{ t-1 }+...+{ 2013 }^{ 2^{ k } }+1 \right]$

Since $t$ is odd, ${ 2013 }^{ n }-1\vdots { 2 }^{ 2014 }\Longleftrightarrow { { 2013 }^{ 2 } }^{ k }-1\vdots { 2 }^{ 2014 }$

Also, ${ { 2013 }^{ 2 } }^{ k }-1=\left( { 2013 }^{ 2 }-1 \right) \left( { 2013 }^{ 2 }+1 \right) \left( { 2013 }^{ 4 }+1 \right) ...\left( { { 2013 }^{ 2 } }^{ k-1 }+1 \right)$

Since $2013\equiv 1\left( mod4 \right) \Longrightarrow { { 2013 }^{ 2 } }^{ i }+1\equiv 2\left( mod4 \right)$

Hence ${ { 2013 }^{ n } }\vdots { 2 }^{ 2014 }\Longleftrightarrow \left( k-1 \right) +3\ge 2014$

Therefore, the minimum value of $n$ is $\boxed { { 2 }^{ 2012 } }$

This problem looks scary, but it's almost immediate once we write it out. Let $2013 = 4k+1$ , where $k$ is odd. We want

$2^{2014} \mid ( 4k+1) ^ n - 1$

By considering the binomial expansion, we have:

$2 ^ { 2014 } \mid (4k)^n + {n \choose n-1 } (4k)^{n-1} + \ldots + {n\choose 2} (4k)^2 + { n \choose 1} (4k)^1$

Observe that since $k$ is odd, if $n$ is odd, then the last term is divisible by 4 not 8, while all of the other terms are divisible by 16. Hence, we must have $n = 2n_1$ .
Again, if $n_1$ is odd, then the last term is divisible by 8 not 16, while all of the other terms are divisible by 32. Hence, $n_1 = 2 n_2$ .
Again, if $n_2$ is odd, then the last term is divisible by 16 not 32, while all of the other terms are divisible by 64. Hence $n_2 = 2 n_3$ .
We can continually repeat this process, until we get "If $n_{2012}$ is odd, then the last term is divisible by $2 ^ {2014}$ (but not $2^{2015}$ ). Thus tells us that we must have $n = 2 ^ { 2012 } \times n_{2012}$ .

Hence, the smallest positive integer is $n = 2 ^ { 2012 }$ .

---

Note: I decided to present a basic approach to this problem. Yes there are shorter approaches which use higher-power theorems like lifting the exponent .