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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 $0\le a\le 9$, $0\le b\le 9$, $0\le c\le 9$, and $n<1000$.

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

Let us check all possible $n$:

- For $n$ with $3$ digits, $n\ge 100$. 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, $n=10a+b$, $n\ge 10$. 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 $a=1$: $1-10+b^2-b=0⟹b^2-b-9=0$ Discriminant: $1+36=37$, not a perfect square.

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

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

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

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

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

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

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

For $a=9$: $81-90+b^2-b=0⟹b^2-b-9=0$ 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: $1,3$. $1^2+3^2=1+9=10\ne 13$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

For $a=9$: $81+b^2=90+b⟹b^2-b=9⟹b^2-b-9=0$ 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:

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