, 2025

Solution 1:

Let the last three digits of $n$
be $abc$. That is, $n$ has units digit $c$, tens digit $b$ and hundreds digit $a$.

(By the end of this solution, we will have demonstrated why considering
only the last three digits of $n$ was
sufficient.)

The units digit of the product $\$2013\times$
n$is equal to the units digit of$$3\times$ c$.$$$$\text{[t] cccccc ! ! !2 ! ! ! ! ! !0 ! ! ! ! ! !1 ! ! ! ! ! !3 ! ! ! ! ! !a ! ! ! ! ! !b ! ! ! ! ! !c ! ! ! ! ! !2 ! ! ! ! ! !0 ! ! ! ! ! !2 ! ! ! ! ! !5 ! ! !}$$$$$$

Since the units digit of the product $2013\times n$ is 5, then the units digit
of $3\times c$ is $5$, and so $c=5$.

(You should confirm for yourself that this is the only possible value of
$c$.) $$\text{[t] cccccc ! !2 ! ! ! ! ! !0 ! ! ! ! ! !1 ! ! ! ! ! !3 ! ! ! ! ! ! ! ! !5 ! ! ! ! ! ! ! ! !1 ! ! ! ! ! ! ! ! ! !0 ! ! ! ! !0 ! ! ! ! ! !6 ! ! ! !5 ! ! !}$$

Continuing the long multiplication, the tens digit of $n$ is $b$, and so the tens digit of $2013\times n$ is equal to the units digit
of $6+3b$, as shown.

Since the units digit of $6+3b$ is
$2$, then the units digit of $3b$ is $6$, and so $b=2$.

(You should confirm for yourself that this is the only possible value of
$b$.) $$\text{[t] cccccc ! ! !2 ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !0 ! ! ! ! ! ! ! ! ! ! ! ! !1 ! ! ! ! ! !3 ! ! ! ! ! !b ! ! ! ! ! !5 ! ! ! ! ! !1 ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !0 ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !0 ! ! ! ! ! ! ! ! ! ! ! !6 ! ! ! ! ! !5 ! ! ! ! ! !3b ! ! ! ! ! !2 ! ! ! ! ! !5 ! ! !}$$

The multiplication completed to this point is shown to the
right.

We have determined that the last two digits of the product $2013\times n$ are $25$ exactly when the last two digits of
$n$ are $25$. $$\text{[t] cccccc ! ! !2 ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !0 ! ! ! ! ! ! ! ! ! ! ! ! !1 ! ! ! ! ! !3 ! ! ! ! ! !2 ! ! ! ! ! !5 ! ! ! ! ! !1 ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !0 ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !0 ! ! ! ! ! ! ! ! ! ! ! ! !6 ! ! ! ! ! !5 ! ! ! ! ! !4 ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !0 ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !2 ! ! ! ! ! ! ! ! ! ! ! !6 ! ! ! ! ! !2 ! ! ! ! ! !5 ! ! !}$$

Continuing the long multiplication, the hundreds digit of $n$ is $a$, and so the hundreds digit of $2013\times n$ is equal to the units digit
of $1+0+2+3a$ (the $1$ is the "carry" from the tens
column).

Since the units digit of $3+3a$ is
$0$, then the units digit of $3a$ is $7$, and so $a=9$.

(You should confirm that this is the only possible value of $a$.) $$\text{[t] cccccc ! ! !2 ! ! ! ! ! ! ! ! ! ! ! ! ! ! !0 ! ! ! ! ! ! ! !1 ! ! ! ! ! !3 ! ! ! ! ! !a ! ! ! ! ! ! !2 ! ! ! ! ! !5 ! ! ! ! ! !1 ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !0 ! ! ! ! ! ! ! ! ! ! ! ! ! !0 ! ! ! ! ! ! !6 ! ! ! ! ! !5 ! ! ! ! ! !4 ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !0 ! ! ! ! ! ! ! ! ! ! ! ! ! !2 ! ! ! ! ! !6 ! ! ! ! ! !3a ! ! ! ! ! !0 ! ! ! ! ! !2 ! ! ! ! ! !5 ! ! !}$$

The last three digits of $\$2013\times$
n$are$025$ exactly when the
last three digits of $n$ are $925$ (that is, $a=9$, $b=2$, $c=5$ are the only possibilities for $a$, $b$, $c$).

The multiplication completed to this point is shown below. $$\text{[t] cccccccc ! ! !2 ! ! ! ! ! !0 ! ! ! ! ! !1 ! ! ! ! ! !3 ! ! ! ! ! !9 ! ! ! ! ! !2 ! ! ! ! ! !5 ! ! ! ! ! !1 ! ! ! ! ! !0 ! ! ! !0 ! ! ! ! !6 ! ! ! ! ! !5 ! ! ! ! ! !4 ! ! ! ! !0 ! ! ! !2 ! ! ! ! !6 ! ! ! ! ! !1 ! ! ! ! !8 ! ! ! ! !1 ! ! ! ! !1 ! ! ! !7 ! ! ! ! ! !1 ! ! ! ! ! ! !8 ! ! ! ! ! ! !6 ! ! ! ! ! ! ! !2 ! ! ! ! ! ! !0 ! ! ! ! ! !2 ! ! ! ! ! !5 ! ! !}$$

This shows that when $n=925$, the
last four digits of the product $\$2013\times$
n$are$2025$, as
required.

Adding additional digits to $n$ will
increase the value of $n$, and since
we are asked for the smallest possible value of $n$, we stop here.

Thus, the smallest possible value of $n$ for which $2013\times n$ has last four digits $2025$,
is $n=925$, and so the sum of the
digits of $n$ is $9+2+5=16$.

Solution 2:

We begin by showing that every positive integer having last two
digits $25$ is a multiple of $25$.

(It is worth noting that it is not true that every multiple of
$25$ has last two digits $25$.)

All positive integers whose last two digits are $25$, are $25$ more than some non-negative multiple
of $100$.

That is, all positive integers whose last two digits are $25$ can be expressed as $100k+25$ for some integer $k\geq0$.

Since $100k$ is divisible by $25$, and $25$ is divisible by $25$, then $100k+25$ is divisible by $25$.

Thus, every positive integer whose last two digits are $25$ is a multiple of $25$, and so $2013\times n$ is a multiple of $25$.

Since $2013=3\times11\times61$ does
not have a prime factor of $5$, then
$2013\times n$ is a multiple of $25$ exactly when $n$ is a multiple of $25$.

The last two digits of $2013\times n$
are equal to the two-digit number formed by the last two digits of the
product of $13$ and the last two
digits of $n$.

What are the last two digits of $n$?
Since $n$ is a multiple of $25$, then the last two digits of $n$ could be $25$, $50$, $75$, or $00$.

(You should confirm that these are the only possibilities.)

Thus, the last two digits of $2013 $\times$
n are equal to the last two digits of $13\times25$$or$$13\times50$$or$$13\times75$$or$$13\times$ 00$, which are$25$,$50$,$75$, and$00$ respectively.

Since we require the last two digits of $2013\times n$ to be $25$, then the last two digits of $n$ are $25$.

We have reduced the problem to finding the smallest value of the
positive integer $n$ with last two
digits $25$ so that the last four
digits of $2013\times n$ are $2025$.

We substitute $n=25, 125, 225, 325,425,
$\dots$$, and so on, in turn, into the product$2013 $\times$ n$.

Evaluating these products, we determine that $2013\times 925 = 1\,862\,025$ is the first
time that the last four digits of $\$2013\times$
n$are$2025$.

Thus, the smallest possible value of $n$ for which $2013\times n$ has last four digits $2025$ is $n=925$, and the sum of the digits of $n$ is $9+2+5=16$.