Let $n$ be a three-digit positive integer such that the last three digits of $n^2$ are the same as those of $n$. This means $n^2 \equiv n \pmod{1000}$.

So, $n^2 - n \equiv 0 \pmod{1000}$, or $n(n-1) \equiv 0 \pmod{1000}$.

Since $n$ and $n-1$ are consecutive integers, they are coprime. Therefore, $1000 = 2^3 \cdot 5^3$ must divide either $n$ or $n-1$.

Case 1: $n \equiv 0 \pmod{1000}$
But $n$ is a three-digit integer, so $100 \leq n \leq 999$, so $n = 1000$ is not possible.

Case 2: $n-1 \equiv 0 \pmod{1000}$
Then $n = 1001$, which is not a three-digit integer.

So we need to consider the factors of $1000$ split between $n$ and $n-1$.

Let $n$ be divisible by $a$ and $n-1$ by $b$, where $a \cdot b = 1000$ and $\gcd(a, b) = 1$.

All pairs $(a, b)$ where $a$ and $b$ are coprime and $a \cdot b = 1000$:
The divisors of $1000$ are $1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000$.

We list all pairs $(a, b)$ such that $a \cdot b = 1000$ and $\gcd(a, b) = 1$:
- $(1, 1000)$
- $(8, 125)$
- $(125, 8)$
- $(5, 200)$
- $(200, 5)$
- $(25, 40)$
- $(40, 25)$

Now, for each pair, $n$ is divisible by $a$, $n-1$ by $b$.
So $n \equiv 0 \pmod{a}$ and $n \equiv 1 \pmod{b}$.

By the Chinese Remainder Theorem, since $a$ and $b$ are coprime, there is a unique solution modulo $1000$ for each pair.
Let us compute $n$ for each pair and check if $n$ is a three-digit integer.

1. $(a, b) = (8, 125)$
$n \equiv 0 \pmod{8}$
$n \equiv 1 \pmod{125}$
Let $n = 8k$, so $8k \equiv 1 \pmod{125}$
Find $k$ such that $8k \equiv 1 \pmod{125}$
The inverse of $8$ modulo $125$ is $47$, since $8 \times 47 = 376 = 1 + 3 \times 125 = 1 \pmod{125}$
So $k \equiv 47 \pmod{125}$
So $n = 8k = 8 \times 47 = 376$
Check if $n$ is three-digit: $376$ is three-digit.

Also, $k$ can be $47 + 125m$, so $n = 8 \times (47 + 125m) = 376 + 1000m$
But $n$ must be three-digit, so only $m = 0$ works.
So $n = 376$.

2. $(a, b) = (125, 8)$
$n \equiv 0 \pmod{125}$
$n \equiv 1 \pmod{8}$
Let $n = 125k$, so $125k \equiv 1 \pmod{8}$
$125 \equiv 5 \pmod{8}$, so $5k \equiv 1 \pmod{8}$
The inverse of $5$ modulo $8$ is $5$, since $5 \times 5 = 25 \equiv 1 \pmod{8}$
So $k \equiv 5 \pmod{8}$
So $n = 125 \times 5 = 625$
Again, $n = 625$ is three-digit.

3. $(a, b) = (5, 200)$
$n \equiv 0 \pmod{5}$
$n \equiv 1 \pmod{200}$
Let $n = 5k$, so $5k \equiv 1 \pmod{200}$
The inverse of $5$ modulo $200$ is $41$, since $5 \times 41 = 205 = 1 + 1 \times 200 = 5 \pmod{200}$
Wait, $5 \times 41 = 205 \equiv 5 \pmod{200}$, not $1$.
Try $5k \equiv 1 \pmod{200}$
Try $k = 1$, $5 \times 1 = 5$; $5 \times 41 = 205$; $205 - 200 = 5$.
Try $k = 41$, $5 \times 41 = 205 \equiv 5 \pmod{200}$
Try $k = 81$, $5 \times 81 = 405 \equiv 5 \pmod{200}$
Try $k = 161$, $5 \times 161 = 805 \equiv 5 \pmod{200}$
So $5k$ always ends with $5$ modulo $200$.
But $5k \equiv 1 \pmod{200}$, so $5k = 1 + 200m$, $k = (1 + 200m)/5$, but $1 + 200m$ must be divisible by $5$.
$1 + 200m \equiv 0 \pmod{5}$
$200m \equiv -1 \pmod{5}$
$200 \equiv 0 \pmod{5}$, so $0 \equiv -1 \pmod{5}$, which is impossible.
So no solution for this pair.

4. $(a, b) = (200, 5)$
$n \equiv 0 \pmod{200}$
$n \equiv 1 \pmod{5}$
Let $n = 200k$, $200k \equiv 1 \pmod{5}$
$200 \equiv 0 \pmod{5}$, so $0 \times k \equiv 1 \pmod{5}$, which is impossible.
So no solution for this pair.

5. $(a, b) = (25, 40)$
$n \equiv 0 \pmod{25}$
$n \equiv 1 \pmod{40}$
Let $n = 25k$, $25k \equiv 1 \pmod{40}$
$25 \equiv 25 \pmod{40}$, so $25k \equiv 1 \pmod{40}$
Find inverse of $25$ modulo $40$.
Try $25 \times 9 = 225 \equiv 25 \pmod{40}$
Try $25 \times 9 = 225$, $225 - 5 \times 40 = 225 - 200 = 25$
Try $25 \times 33 = 825$, $825 - 20 \times 40 = 825 - 800 = 25$
Try $25 \times 33 = 825 \equiv 25 \pmod{40}$
Try $25 \times 33 = 825$, $825 - 20 \times 40 = 825 - 800 = 25$
Try $k = 1$, $25 \times 1 = 25$
Try $k = 25$, $25 \times 25 = 625$, $625 - 15 \times 40 = 625 - 600 = 25$
Try $k = 17$, $25 \times 17 = 425$, $425 - 10 \times 40 = 425 - 400 = 25$
So $25k$ always ends with $25$ modulo $40$.
So $25k \equiv 1 \pmod{40}$, $25k = 1 + 40m$, $k = (1 + 40m)/25$, $1 + 40m \equiv 0 \pmod{25}$
$40m \equiv -1 \pmod{25}$
$40 \equiv 15 \pmod{25}$, so $15m \equiv -1 \pmod{25}$
Find $m$ such that $15m \equiv 24 \pmod{25}$
Try $m = 24$, $15 \times 24 = 360$, $360 \equiv 10 \pmod{25}$
Try $m = 16$, $15 \times 16 = 240$, $240 \equiv 15 \pmod{25}$
Try $m = 9$, $15 \times 9 = 135$, $135 \equiv 10 \pmod{25}$
Try $m = 19$, $15 \times 19 = 285$, $285 \equiv 10 \pmod{25}$
Try $m = 14$, $15 \times 14 = 210$, $210 \equiv 10 \pmod{25}$
Try $m = 21$, $15 \times 21 = 315$, $315 \equiv 15 \pmod{25}$
Try $m = 23$, $15 \times 23 = 345$, $345 \equiv 20 \pmod{25}$
Try $m = 24$, $15 \times 24 = 360$, $360 \equiv 10 \pmod{25}$
Try $m = 1$, $15 \times 1 = 15$, $15 \equiv 15 \pmod{25}$
Try $m = 2$, $15 \times 2 = 30$, $30 \equiv 5 \pmod{25}$
Try $m = 3$, $15 \times 3 = 45$, $45 \equiv 20 \pmod{25}$
Try $m = 4$, $15 \times 4 = 60$, $60 \equiv 10 \pmod{25}$
Try $m = 5$, $15 \times 5 = 75$, $75 \equiv 0 \pmod{25}$
So $m = 5$, $15 \times 5 = 75 \equiv 0 \pmod{25}$, so $15m \equiv 0 \pmod{25}$
But we want $15m \equiv 24 \pmod{25}$
Try $m = 24$, $15 \times 24 = 360 \equiv 10 \pmod{25}$
Try $m = 9$, $15 \times 9 = 135 \equiv 10 \pmod{25}$
Try $m = 11$, $15 \times 11 = 165 \equiv 15 \pmod{25}$
Try $m = 16$, $15 \times 16 = 240 \equiv 15 \pmod{25}$
Try $m = 23$, $15 \times 23 = 345 \equiv 20 \pmod{25}$
Try $m = 24$, $15 \times 24 = 360 \equiv 10 \pmod{25}$
Try $m = 17$, $15 \times 17 = 255 \equiv 5 \pmod{25}$
Try $m = 8$, $15 \times 8 = 120 \equiv 20 \pmod{25}$
Try $m = 19$, $15 \times 19 = 285 \equiv 10 \pmod{25}$
Try $m = 21$, $15 \times 21 = 315 \equiv 15 \pmod{25}$
Try $m = 22$, $15 \times 22 = 330 \equiv 5 \pmod{25}$
Try $m = 13$, $15 \times 13 = 195 \equiv 20 \pmod{25}$
Try $m = 18$, $15 \times 18 = 270 \equiv 20 \pmod{25}$
Try $m = 6$, $15 \times 6 = 90 \equiv 15 \pmod{25}$
Try $m = 7$, $15 \times 7 = 105 \equiv 5 \pmod{25}$
Try $m = 12$, $15 \times 12 = 180 \equiv 5 \pmod{25}$
Try $m = 20$, $15 \times 20 = 300 \equiv 0 \pmod{25}$
Try $m = 25$, $15 \times 25 = 375 \equiv 0 \pmod{25}$
So, perhaps this is not working. Let's try another approach.
Alternatively, since $25k \equiv 1 \pmod{40}$, $25k = 1 + 40m$, $k = (1 + 40m)/25$, $1 + 40m$ must be divisible by $25$.
$1 + 40m \equiv 0 \pmod{25}$
$40m \equiv -1 \pmod{25}$
$40 \equiv 15 \pmod{25}$, so $15m \equiv 24 \pmod{25}$
Now, $15m \equiv 24 \pmod{25}$
The inverse of $15$ modulo $25$ is $5$, since $15 \times 5 = 75 \equiv 0 \pmod{25}$, so $15 \times 2 = 30 \equiv 5 \pmod{25}$, $15 \times 7 = 105 \equiv 5 \pmod{25}$
Try $m = 24$, $15 \times 24 = 360 \equiv 10 \pmod{25}$
Try $m = 19$, $15 \times 19 = 285 \equiv 10 \pmod{25}$
Try $m = 9$, $15 \times 9 = 135 \equiv 10 \pmod{25}$
Try $m = 16$, $15 \times 16 = 240 \equiv 15 \pmod{25}$
Try $m = 21$, $15 \times 21 = 315 \equiv 15 \pmod{25}$
Try $m = 23$, $15 \times 23 = 345 \equiv 20 \pmod{25}$
Try $m = 8$, $15 \times 8 = 120 \equiv 20 \pmod{25}$
Try $m = 13$, $15 \times 13 = 195 \equiv 20 \pmod{25}$
Try $m = 22$, $15 \times 22 = 330 \equiv 5 \pmod{25}$
Try $m = 12$, $15 \times 12 = 180 \equiv 5 \pmod{25}$
Try $m = 17$, $15 \times 17 = 255 \equiv 5 \pmod{25}$
Try $m = 14$, $15 \times 14 = 210 \equiv 10 \pmod{25}$
Try $m = 11$, $15 \times 11 = 165 \equiv 15 \pmod{25}$
Try $m = 6$, $15 \times 6 = 90 \equiv 15 \pmod{25}$
Try $m = 4$, $15 \times 4 = 60 \equiv 10 \pmod{25}$
Try $m = 3$, $15 \times 3 = 45 \equiv 20 \pmod{25}$
Try $m = 2$, $15 \times 2 = 30 \equiv 5 \pmod{25}$
Try $m = 1$, $15 \times 1 = 15 \equiv 15 \pmod{25}$
Try $m = 0$, $15 \times 0 = 0 \equiv 0 \pmod{25}$
So, perhaps there is no solution for this pair.

6. $(a, b) = (40, 25)$
$n \equiv 0 \pmod{40}$
$n \equiv 1 \pmod{25}$
Let $n = 40k$, $40k \equiv 1 \pmod{25}$
$40 \equiv 15 \pmod{25}$, so $15k \equiv 1 \pmod{25}$
The inverse of $15$ modulo $25$ is $12$, since $15 \times 12 = 180 = 7 \times 25 + 5 = 5 \pmod{25}$
Wait, $15 \times 12 = 180$, $180 - 7 \times 25 = 180 - 175 = 5$
Try $k = 2$, $15 \times 2 = 30 \equiv 5 \pmod{25}$
Try $k = 7$, $15 \times 7 = 105 \equiv 5 \pmod{25}$
Try $k = 17$, $15 \times 17 = 255 \equiv 5 \pmod{25}$
Try $k = 23$, $15 \times 23 = 345 \equiv 20 \pmod{25}$
Try $k = 24$, $15 \times 24 = 360 \equiv 10 \pmod{25}$
Try $k = 9$, $15 \times 9 = 135 \equiv 10 \pmod{25}$
Try $k = 11$, $15 \times 11 = 165 \equiv 15 \pmod{25}$
Try $k = 1$, $15 \times 1 = 15 \equiv 15 \pmod{25}$
Try $k = 3$, $15 \times 3 = 45 \equiv 20 \pmod{25}$
Try $k = 8$, $15 \times 8 = 120 \equiv 20 \pmod{25}$
Try $k = 13$, $15 \times 13 = 195 \equiv 20 \pmod{25}$
Try $k = 18$, $15 \times 18 = 270 \equiv 20 \pmod{25}$
Try $k = 19$, $15 \times 19 = 285 \equiv 10 \pmod{25}$
Try $k = 14$, $15 \times 14 = 210 \equiv 10 \pmod{25}$
Try $k = 16$, $15 \times 16 = 240 \equiv 15 \pmod{25}$
Try $k = 21$, $15 \times 21 = 315 \equiv 15 \pmod{25}$
Try $k = 22$, $15 \times 22 = 330 \equiv 5 \pmod{25}$
Try $k = 12$, $15 \times 12 = 180 \equiv 5 \pmod{25}$
Try $k = 5$, $15 \times 5 = 75 \equiv 0 \pmod{25}$
Try $k = 10$, $15 \times 10 = 150 \equiv 0 \pmod{25}$
Try $k = 15$, $15 \times 15 = 225 \equiv 0 \pmod{25}$
Try $k = 20$, $15 \times 20 = 300 \equiv 0 \pmod{25}$
So, perhaps there is no solution for this pair.

Therefore, the only three-digit solutions are $n = 376$ and $n = 625$.

Check:
$376^2 = 141376$, last three digits are $376$.
$625^2 = 390625$, last three digits are $625$.

Thus, the answer is $\boxed{376}$ and $\boxed{625}$.