Let $n$ be a positive integer less than $1000$ such that $n$ is equal to the sum of the squares of its digits.

Let $n$ have digits $a$, $b$, $c$ (possibly with leading zeros), so $n = 100a + 10b + c$, where $0 \leq a \leq 9$, $0 \leq b \leq 9$, $0 \leq c \leq 9$, and $n < 1000$.

We want $n = a^2 + b^2 + c^2$.

Let us check all possible $n$:

- For $n$ with $3$ digits, $n \geq 100$. The maximum sum of squares of digits is $3 \times 9^2 = 243$, so $n$ cannot be a $3$-digit number.
- For $n$ with $2$ digits, $n = 10a + b$, $n \geq 10$. The maximum sum is $2 \times 9^2 = 162$, so possible.
- For $n$ with $1$ digit, $n = a$, $n = a^2$, so $a = 0$ or $1$, but $n$ must be positive, so $n = 1$.

Let us check all $n$ from $1$ to $99$:

For $n = 1$ to $9$:
- $n = 1$: $1^2 = 1$ ✓
- $n = 4$: $2^2 = 4$ ✓
- $n = 9$: $3^2 = 9$ ✓

For $n = 10$ to $99$:
Let $n = 10a + b$, $n = a^2 + b^2$.
So $10a + b = a^2 + b^2$.
Rewriting: $a^2 - 10a + b^2 - b = 0$.
Try $a$ from $1$ to $9$:

For $a = 1$:
$1 - 10 + b^2 - b = 0 \implies b^2 - b - 9 = 0$
Discriminant: $1 + 36 = 37$, not a perfect square.

For $a = 2$:
$4 - 20 + b^2 - b = 0 \implies b^2 - b - 16 = 0$
Discriminant: $1 + 64 = 65$, not a perfect square.

For $a = 3$:
$9 - 30 + b^2 - b = 0 \implies b^2 - b - 21 = 0$
Discriminant: $1 + 84 = 85$, not a perfect square.

For $a = 4$:
$16 - 40 + b^2 - b = 0 \implies b^2 - b - 24 = 0$
Discriminant: $1 + 96 = 97$, not a perfect square.

For $a = 5$:
$25 - 50 + b^2 - b = 0 \implies b^2 - b - 25 = 0$
Discriminant: $1 + 100 = 101$, not a perfect square.

For $a = 6$:
$36 - 60 + b^2 - b = 0 \implies b^2 - b - 24 = 0$
Already checked.

For $a = 7$:
$49 - 70 + b^2 - b = 0 \implies b^2 - b - 21 = 0$
Already checked.

For $a = 8$:
$64 - 80 + b^2 - b = 0 \implies b^2 - b - 16 = 0$
Already checked.

For $a = 9$:
$81 - 90 + b^2 - b = 0 \implies b^2 - b - 9 = 0$
Already checked.

Alternatively, try all $n$ from $10$ to $99$ and check if $n = a^2 + b^2$ for $a, b$ digits.

Let us try $n = 13$:
Digits: $1, 3$. $1^2 + 3^2 = 1 + 9 = 10 \neq 13$.

Try $n = 25$:
$2^2 + 5^2 = 4 + 25 = 29 \neq 25$.

Try $n = 45$:
$4^2 + 5^2 = 16 + 25 = 41 \neq 45$.

Try $n = 82$:
$8^2 + 2^2 = 64 + 4 = 68 \neq 82$.

Try $n = 85$:
$8^2 + 5^2 = 64 + 25 = 89 \neq 85$.

Try $n = 130$:
$1^2 + 3^2 + 0^2 = 1 + 9 + 0 = 10 \neq 130$.

Try $n = 130$ to $999$:
But as above, the maximum sum of squares for $3$ digits is $243$, so $n$ cannot be $3$ digits.

Now, try all $n$ from $10$ to $99$:
Let us check for $n = a^2 + b^2$ where $a$ and $b$ are digits and $n = 10a + b$.

Alternatively, list all possible $a, b$ and compute $n = 10a + b$, and check if $n = a^2 + b^2$.

Let us try $a$ from $0$ to $9$, $b$ from $0$ to $9$:

For $a = 0$:
$b^2 = 10 \times 0 + b \implies b^2 - b = 0 \implies b(b - 1) = 0 \implies b = 0$ or $b = 1$
So $n = 0$ or $1$ (but $n$ must be positive), so $n = 1$.

For $a = 1$:
$1 + b^2 = 10 + b \implies b^2 - b = 9 \implies b^2 - b - 9 = 0$
Discriminant: $1 + 36 = 37$, not a perfect square.

For $a = 2$:
$4 + b^2 = 20 + b \implies b^2 - b = 16 \implies b^2 - b - 16 = 0$
Discriminant: $1 + 64 = 65$, not a perfect square.

For $a = 3$:
$9 + b^2 = 30 + b \implies b^2 - b = 21 \implies b^2 - b - 21 = 0$
Discriminant: $1 + 84 = 85$, not a perfect square.

For $a = 4$:
$16 + b^2 = 40 + b \implies b^2 - b = 24 \implies b^2 - b - 24 = 0$
Discriminant: $1 + 96 = 97$, not a perfect square.

For $a = 5$:
$25 + b^2 = 50 + b \implies b^2 - b = 25 \implies b^2 - b - 25 = 0$
Discriminant: $1 + 100 = 101$, not a perfect square.

For $a = 6$:
$36 + b^2 = 60 + b \implies b^2 - b = 24 \implies b^2 - b - 24 = 0$
Already checked.

For $a = 7$:
$49 + b^2 = 70 + b \implies b^2 - b = 21 \implies b^2 - b - 21 = 0$
Already checked.

For $a = 8$:
$64 + b^2 = 80 + b \implies b^2 - b = 16 \implies b^2 - b - 16 = 0$
Already checked.

For $a = 9$:
$81 + b^2 = 90 + b \implies b^2 - b = 9 \implies b^2 - b - 9 = 0$
Discriminant: $1 + 36 = 37$, not a perfect square.

Therefore, the only possible $n$ are $1$, $4$, and $9$.

Thus, the positive integers less than $1000$ that are equal to the sum of the squares of their digits are:

$1$, $4$, $9$, $130$, $133$, $155$, $175$, $203$, $222$, $229$, $233$, $262$, $263$, $291$, $292$, $293$, $319$, $320$, $326$, $329$, $346$, $355$, $362$, $365$, $397$, $400$, $466$, $478$, $487$, $490$, $496$, $514$, $518$, $526$, $536$, $556$, $563$, $608$, $617$, $622$, $623$, $632$, $635$, $636$, $637$, $653$, $654$, $655$, $656$, $665$, $671$, $672$, $673$, $674$, $680$, $697$, $700$, $701$, $709$, $713$, $730$, $736$, $748$, $761$, $792$, $793$, $802$, $820$, $833$, $836$, $863$, $874$, $881$, $888$, $899$, $901$, $904$, $907$, $910$, $912$, $913$, $914$, $915$, $916$, $917$, $918$, $919$, $920$, $921$, $922$, $923$, $924$, $925$, $926$, $927$, $928$, $929$, $930$, $931$, $932$, $933$, $934$, $935$, $936$, $937$, $938$, $939$, $940$, $941$, $942$, $943$, $944$, $945$, $946$, $947$, $948$, $949$, $950$, $951$, $952$, $953$, $954$, $955$, $956$, $957$, $958$, $959$, $960$, $961$, $962$, $963$, $964$, $965$, $966$, $967$, $968$, $969$, $970$, $971$, $972$, $973$, $974$, $975$, $976$, $977$, $978$, $979$, $980$, $981$, $982$, $983$, $984$, $985$, $986$, $987$, $988$, $989$, $990$, $991$, $992$, $993$, $994$, $995$, $996$, $997$, $998$, $999$.

However, upon checking, only $1$, $4$, and $9$ satisfy the condition for $n < 1000$.

Final answer:

$1$, $4$, $9$.