I hope so. But please double check (or, better, simplify) the argument below.
Denote $N=qs^2$ for $q$ squarefree. Then each $d_i$ divides $s$, say $d_i=s/m_i$ and we get $$q=1/s^2+\sum_{i=1}^r 1/m_i^2, \quad\quad\quad (\heartsuit)$$
and the sum of $r+1$ square reciprocals is integer. Since $d_i\geqslant 3$, we get $m_i\leqslant s/3$. Subtracting the summands with $m_i=1$ we get $(\heartsuit)$ with smaller $r$ and the same condition $m_i\leqslant s/3$ (possibly $q$ is no longer squarefree, but we do not care anymore.) So, now each $m_i$ is at least 2. So, we assume that $(\heartsuit)$ holds with $r\leqslant 8$, some integer $q$ and $s/3\geqslant m_i\geqslant 2$.
If $r=8$ and all $m_i$ are equal to 2, then $1/s^2$ must be an integer which is absurd. Otherwise $\sum 1/m_i^2\leqslant 7\cdot 1/4+1/9$ and $1/s^2+\sum 1/m_i^2\leqslant 1/36+7/4+1/9<2$, thus $q=1$.
Further we denote $s=m_{r+1}$, and $(\heartsuit)$ reads as $$1=\sum_{i=1}^{r+1} 1/m_i^2. \quad\quad\quad(\clubsuit)$$
And we have a
Special Condition. $m_{r+1}\geqslant 3\max(m_1,\ldots,m_r)$ and $m_i$ divides $m_{r+1}$ for all $i$.
What we do below is finding all solutions of $(\clubsuit)$ with $r\leqslant 8$. They all fail Special Condition. Let me start with listing them:
(a) $r=3$, $(2,2,2,2)$;
(b) $r=8$, $(2,2,2,3,6,6,6,6,6)$;
(c) $r=8$, $(2,2,3,3,4,4,4,4,6)$;
(d) $r=8$, $(2,2,2,4,4,4,6,6,12)$;
(e) $r=6$, $(2,2,2,4,4,4,4)$;
(f) $r=8$, $(2,2,2,3,3,12,12,12,12)$;
(g) $r=7$, $(2,2,2,3,4,4,12,12)$;
(h) $r=8$, $(2,2,2,3,9,4,4,36,36)$;
(i) $r=8$, $(2,2,2,3,4,4,12,15,20)$;
(j) $r=7$, $(2,2,3,3,3,3,6,6)$;
(k) $r=5$, $(2,2,2,3,3,6)$;
(l) $r=7$, $(2,2,2,3,3,7,14,21)$;
(m) $r=7$, $(2,2,2,3,3,9,9,18)$;
(n) $r=8$, $(3,3,3,3,3,3,3,3,3)$
(o) $r=0$, $(1)$.
Now goes a rather lengthy and boring proof that there are no other solutions of $(\clubsuit)$.
Denote by $2^\beta$ the maximal power of 2 which divides one of the numbers $m_i$, $i=1,\ldots,r+1$. We have $\beta\geqslant 1$: otherwise all $1/m_i^2$ are at most $1/9$, and not all equal to $1/9$, and RHS of $(\clubsuit)$ is less than 1. If we have exactly $h$ indices $i$ for which $2^\beta$ divides $m_i$, then multiplying $(\clubsuit)$ by $2^{2\beta}$ and considering the expression modulo 4 we get $4|h$. Thus $h=8$ or $h=4$ or $h=0$.
Consider several cases.
$h=0$, all $m_i$'s are odd. Then either $r=0$ and we get solution (0), or they all are not less than 3, and $\sum 1/m_i^2\leqslant 9\cdot 1/9=1$, with equality corresponding to solution (n).
$h=8$ and $\beta \geqslant 2$. Then RHS of $(\clubsuit)$ does not exceed $8\cdot \frac1{16}+1\cdot \frac14<1$.
$h=8$ and $\beta=1$. Then eight even $m_i$'s may be denoted by $2s_i$, $s_i$ are odd, $i=1,\ldots,8$, and $(\clubsuit)$ reads as
$$
\frac14\sum_{i=1}^8\frac1{s_i^2}+\frac{\varepsilon}{\ell^2}=1,\quad\quad\quad(\smile)
$$
where $\varepsilon=0$ if $r=7$ and $\varepsilon=1$ if $r=8$ (and $\ell$ is odd). At most three $s_i$'s may be equal to 1, others are at least 3, also $\ell\geqslant 3$, thus LHS of $(\smile)$ is at most
$\frac14(3\cdot1+5\cdot \frac19)+\frac19=1$, the equality case is unique, it is the solution (b).
$h=4$ and $\beta\geqslant 3$. Analogously, $(\clubsuit)$ reads as
$$
\frac1{4^\beta}\sum_{i=1}^4 \frac1{s_i^2}+\sum_{i=1}^{r-3}\frac1{\ell_i^2}=1,\quad\quad\quad(\ast)
$$
where $s_i$ are odd. At most 3 $\ell_i$'s are equal to 2. Assume that at least one of $\ell_i$'s is at least 4. Then LHS of $(\ast)$ is at most
$\frac1{64}\cdot 4+3\cdot\frac14+1\cdot\frac19+1\cdot\frac1{16}<1$, a contradiction. So, all $\ell_i$'s are equal to 2 or 3, and we get contradiction modulo 8 after multiplying by $4^\beta$.
$h=4$, $\beta=2$. Now we get
$$
\frac1{16}\sum_{i=1}^4 \frac1{s_i^2}+\sum_{i=1}^{r-3}\frac1{\ell_i^2}=1,\quad\quad\quad(\diamond)
$$
where $s_i$ are odd and $\ell_i$ not divisible by 4. Assume that at most one $\ell_i$'s is equal to 2, then LHS of $(\diamond)$ does not exceed
$\frac1{16}\cdot 4+1\cdot \frac14+4\cdot \frac19<1$, a contradiction. Thus we get $r\geqslant 5$ and we may suppose $\ell_{r-3}=\ell_{r-4}=2$, and $(\diamond)$ reads as
$$
\frac1{16}\sum_{i=1}^4 \frac1{s_i^2}+\sum_{i=1}^{r-5}\frac1{\ell_i^2}=\frac12.\quad\quad\quad(\diamond\diamond)
$$
If all $\ell_i$'s ($i=1,\ldots,r-5$) are odd, we get a contradiction modulo 8 after multiplying by $16$. If none of $\ell_i$'s ($i=1,\ldots,r-5$) equals to 2, then one of them is at least 6, and LHS of $(\diamond\diamond)$ does not exceed $\frac1{16}\cdot 4+2\cdot \frac19+\frac1{36}=1/2$. There exists unique equality case, it corresponds to solution (c). Thus, we may suppose that $r\geqslant 6$ and $\ell_{r-5}=2$, and $(\diamond\diamond)$ reads as
$$
\frac1{16}\sum_{i=1}^4 \frac1{s_i^2}+\sum_{i=1}^{r-6}\frac1{\ell_i^2}=\frac14.\quad\quad\quad(\diamond\diamond\diamond)
$$
If exactly one of $\ell_i$'s ($i=1,\ldots,r-6$) is even (recall that it is not however divisible by 4), we get a contradiction modulo 8 after multiplying by 16. If two of $\ell_i$'s ($i=1,\ldots,r-6$) are even, then they are at least 6, and LHS of $(\diamond\diamond\diamond)$ is at most
$\frac1{16}(3+\frac19)+2\cdot \frac1{36}=\frac14$. Again, the equality case is unique, it provides a solution (d). Thus, all $\ell_i$'s ($i=1,\ldots,r-6$) are odd. If $r=6$, we immediately get $s_1=\ldots=s_4=1$, this gives solution (e). So, $r\geqslant 7$, at most 3 of $s_i$'s are equal to 1 and $$\sum_{i=1}^{r-6}\frac1{\ell_i^2}\geqslant \frac14-\frac1{16}\left(3+\frac19\right)=\frac1{18}.$$
Thus there is 3 or 5 between $\ell_i$'s, $1\leqslant i\leqslant r-6$.
Well, proceed with cases. Assume that $\ell_{r-6}=5$, but $\ell_{r-7}\ne 3$ (or $r=7$ and $\ell_1=5$). Then $\sum \frac1{s_i^2}=4-16\sum_{i=1}^{r-6}\frac1{\ell_i^2}\geqslant 4-\frac{32}{25}$. Therefore at least three $s_i$'s are equal to 1, say $s_2=s_3=s_4=1$, and $(\diamond\diamond\diamond)$ reads as $\frac1{16s_1^2}+\frac{\varepsilon}{\ell_{r-7}^2}=\frac1{16}-\frac1{25}=\frac9{16\cdot 25}$ (here $\varepsilon=0$ if $r=7$ and $\varepsilon=1$ if $r=8$). We see that $r=8$, and we get an equation $\frac1{\ell_1^2}=\frac{9}{16\cdot 25}-\frac1{16s_1^2}=\frac{(3s_1-5)(3s_1+5)}{16\cdot 25\cdot s_1^2}$. If 5 does not divide $s_1$, then $3s_1\pm 5$ is coprime to $25s_1^2$, and the numerator of $\frac{(3s_1-5)(3s_1+5)}{16\cdot 25\cdot s_1^2}$ may be equal to 1 only if both $3s_1-5$ and $3s_1+5$ are powers of 2, which is impossible (powers of 2 never differ by 10). So, $s_1$ is divisible by 5 and $\frac9{16\cdot 25}\geqslant \frac1{\ell_1^2}\geqslant \frac8{16\cdot 25}$, that is, $\ell_1=7$, but this does not provide a solution.
Assume then that $\ell_{r-6}=3$. Analogously, we get $\sum \frac1{s_i^2}+\frac{16\varepsilon}{\ell_{r-7}^2}=16(\frac14-\frac19)=\frac{20}{9}$. If all $s_i$'s are at least 3, then LHS is at most $4/9+16/9=20/9$, and equality holds in the unique case, corresponding to solution $(f)$.
So let us suppose that $s_4=1$, we get $\sum_{i=1}^3 \frac1{s_i^2}+\frac{16\varepsilon}{\ell_{r-7}^2}=\frac{11}9$. If all $s_1,s_2,s_3$ are at least 3, we get $r=8$ and $\frac{11}9\geqslant \frac{16}{\ell_1^2}\geqslant \frac{11}9-3\cdot \frac19=\frac{8}9$, that is impossible (recall that $\ell_1$ is odd). So, without loss of generality $s_3=1$ and $\frac1{s_1^2}+\frac1{s_2^2}+\frac{16\varepsilon}{\ell_1^2}=\frac29$. Clearly $s_1,s_2\geqslant 3$. If $r=7$, i.e., $\varepsilon=0$, this yields a solution with $s_1=s_2=3$, that is (g). So assume that $r=8$. Since $\frac{16}{\ell_1^2}<\frac2{9}$, we get $\ell_1\geqslant 9$. Thus $\frac1{s_1^2}+\frac1{s_2^2}\geqslant \frac29-\frac{16}{81}=\frac2{81}$. It yields $\min(s_1,s_2)\leqslant 9$.
If now $s_1,s_2$ are at least 5, then $\frac{16}{\ell_1^2}\geqslant \frac29-\frac2{25 }=\frac{32}{225}$, or $\frac{225}2\geqslant \ell_1^2$, this yields $\ell_1=9$. Thus $\frac1{s_1^2}+\frac1{s_2^2}=\frac2{81}$. If $s_1=s_2=9$, we get solution (h). Otherwise, say, $s_2=7$,
this does not lead to a solution.
Finally, if $s_2=3$, we get an equation $\frac1{s_1^2}+\frac{16}{\ell_1^2}=\frac19$, or
$\frac{16}{\ell_1^2}=\frac19-\frac1{s_1^2}=\frac{(s_1-3)(s_1+3)}{9s_1^2}$. If 3 does not divide $s_1$, then $s_1\pm 3$ are coprime to $9s_1^2$, so both $s_1-3$ and $s_1+3$ must be powers of 2. This happens only for $s_1=5$, and we get solution
(i). If $s_1=3k$, we get $\frac{(k-1)(k+1)}{9k^2}=\frac{16}{s_1^2}$, thus $k^2-1=(k-1)(k+1)$ is a positive perfect square, which is impossible.
- Uff, now let $h=4$ and $\beta=1$. We get
$$
\frac14\sum_{i=1}^4\frac1{s_i^2}+\sum_{i=1}^{r-3}\frac1{\ell_i^2}=1
$$
for odd $s_i$ and $\ell_i$. Multiplying by 4 and considering modulo 8 we see that $r-3$ must be even, thus $r$ is odd, so $r\leqslant 7$. Therefore $\sum\frac1{\ell_i^2}\leqslant \frac49$ and $\sum\frac1{s_i^2}\geqslant 4(1-\frac49)=\frac{20}9$. This implies that at least two of $s_i$'s are equal to 1, say $s_3=s_4=1$. We get from above $\frac1{s_1^2}+\frac1{s_2^2}\geqslant \frac29$, so either $r=7$, $\ell_1=\ldots=\ell_4=3$, $s_1=s_2=3$,
that's solution (j), or, say, $s_2=1$, and we get $\frac1{4s_1^2}+\sum \frac1{\ell_i^2}=\frac14$. There is a solution (a) with $s_1=1$ and $r=3$. Otherwise $s_1\geqslant 3$ and we get $\sum \frac1{\ell_i^2}\in (\frac14-\frac1{36},\frac14)$. Thus, one of $\ell_i$'s is equal to 3, say $\ell_{r-3}=1$, and $\frac1{4s_1^2}+\sum_{i=1}^{r-4}\frac1{\ell_i^2}=\frac5{36}$. Assume that all remaining in LHS $\ell_i$'s are greater than 3, and that $s_1$ is greater than 3. Then $\frac1{4s_1^2}+\sum_{i=1}^{r-4}\frac1{\ell_i^2}\leqslant \frac1{100}+3\cdot \frac1{25}=\frac{13}{100}<\frac5{36}$.
Next case. $s_1=3$. Then $\sum_{i=1}^{r-4}\frac1{\ell_i^2}=\frac19$. If $r=5$, we get solution (k). Otherwise $r\geqslant 6$, then remaining $\ell_i$'s are at least 5, not all equal to 5, thus $\sum_{i=1}^{r-4}\frac1{\ell_i^2}\leqslant \frac1{25}+\frac1{25}+\frac1{49}<\frac19$.
Finally, let $s_1>3$, but $\ell_{r-4}=3$. Then $\frac1{4s_1^2}+\sum_{i=1}^{r-5}\frac1{\ell_i^2}=\frac1{36}$. If all $\ell_i$'s, $i\leqslant r-5$, are at least 11, then LHS is at most $1/100+2/121<1/36$. So, one of them equals to 7 or 9.
This leads to equations $\frac1{4s_1^2}+\frac1{\ell_1^2}=\frac{13}{36\cdot 49}$ and $\frac1{4s_1^2}+\frac1{\ell_1^2}=\frac{5}{4\cdot 81}$, respectively.
For $\frac1{4s_1^2}+\frac1{\ell_1^2}=\frac{13}{36\cdot 49}$ note that 7 must divide both $s_1$ and $\ell_1$ (otherwise we get a contradiction modulo 7 multiplying by 49, due to 13 being quadratic non-residue modulo 7). If, say, $s_1=7z$, $\ell_1=7w$, we get an equation $\frac1{4z^2}+\frac1{w^2}=\frac{13}{36}=\frac14+\frac19$. This is only possible for $z=1$, $w=3$ (for $w=1$ LHS is too large, otherwise too small). Surprisingly, this solution does not enjoy SC again.
For $\frac1{4s_1^2}+\frac1{\ell_1^2}=\frac{5}{4\cdot 81}$, we analogously get that $s_1$, $\ell_1$ must be divisible by 9 (since 5 is not a quadratic residue modulo 3). So, we get an equation $\frac1{4z^2}+\frac1{w^2}=\frac{5}4$ which has the unique solution $z=w=1$, which leads to solution (m).