Sum of squares and divisibility Consider an integer of the form $$N = 1 + \sum_{i=1}^r d_i^2$$ where $d_i \in \mathbb{N}_{\ge 3}$ and $d_i^2$ divides $N$.
Question: Must $r$ be greater than or equal to $9$?
Checking (with SageMath):   It is true for $N \le 500000$.
Remark: If it is true in general then it is optimal because $$144 = 1 + 3^2 + 3^2 + 3^2+ 4^2  + 4^2  + 4^2  + 4^2  + 4^2  + 6^2$$

Motivation (from tensor category theory): If above question has a positive answer then by [1, Proposition 8.14.6] and [2, Theorem 3.4] a simple integral modular fusion category (over $\mathbb{C}$) would be of rank $ \ge 10 $. And for so, if required, we can also assume $d_i$ not a prime-power by [3, Corollary 6.16]. Then the (SageMath) checking is $N \le 10^6$; and above optimality would be unchanged because $$ 116964 = 1 + 6^2 +  18^2 + 38^2 + 38^2 + 114^2 + 114^2 + 171^2 + 171^2 + 171^2$$
We did not add this additional assumption at the beginning because it seems to be true without it (experimentally), and without it the question is (number theoretically) more interesting.
With this additional assumption, the next example with $r=9$ is $$ 396900 = 1 + 18^2 + 30^2 + 70^2 + 70^2 + 210^2 + 210^2 + 315^2 + 315^2 + 315^2 $$
Above two examples with $r=9$ can be excluded from coming from a fusion ring because:

*

*First one: consider the simple object of FPdim $6$, multiply it with its dual (which must be itself here) and apply FPdim (ring homomorphism). Then you get $6^2=1+\cdots$, but there is no FPdim $\le 35$ (of a non-trivial simple object) which is not a multiple of $6$.

*Second one: as above, $18^2 - 1 = 323$ is odd and the only odd FPdim for a non-trivial simple object is $315$ , but $323-315 = 8$ and there is no non-trivial simple object of FPdim $\le 8$.

Conclusion: if above two examples are the only ones for $r=9$ (I checked that there is no other one for $N \le 10^6$), and if above question has a positive answer (we can put the additional assumption if required), then a simple integral modular fusion category (over $\mathbb{C}$) would be of rank $ \ge 11 $.
References
[1]: P. Etingof, S. Gelaki, D. Nikshych, V. Ostrik; Tensor categories. Mathematical Surveys and Monographs (2015) 205.
[2]: J. Dong, S. Natale, L. Vendramin; Frobenius property for fusion categories of small integral dimension. J. Algebra Appl. 14 (2015), no. 2, 1550011.
[3]: D. Nikshych, Morita equivalence methods in classification of fusion categories, Hopf algebras and tensor categories, 289-325,
Contemp. Math., 585, Amer. Math. Soc. (2013).
 A: The answer of Francesco Polizzi recasts the problem into a form in which known results prove at least that there are (at most) a finite number of exceptions.
For any positive integer $s$, E. Landau proved (around 1909) that there are only a finite number of solutions in positive integers $m_{i}$ to the equation $\sum_{i=1}^{s} \frac{1}{m_{i}} = 1.$ Hence, since in the equation arising in the problem here (and violating the required condition) we have $s \leq 10$, we see that there are only finitely many possibilities for the integer $N$ if $r \leq 9$.
It is possible to get an explicit bound using Landau's theorem, but it would be very large. Perhaps the extra conditions here can reduce it to manageable levels.
A: This is not a complete answer, but just a way to transform the problem into one that can be attacked by brute force in some known way.
Write $d_i^2=N/n_i$. Then your relation becomes $$\frac{1}{n_1}+ \frac{1}{n_2} + \ldots +\frac{1}{n_r} + \frac{1}{N}=1,$$
where we can assume $n_1 \leq n_2 \leq \ldots \leq n_r \leq N$.
So your problem boils down to checking all the possible decompositions of $1$ as sum of Egyptian fractions of length $r+1 \leq 9$, and looking for one such that $N/n_i$  is a square greater than or equal to $9$ (i.e. $3^2$) for all $i \in \{1, \ldots, r\}$.
I do not see at the moment a better way to do this than performing a systematic computer check. In fact, many things are already known. For instance, the number of different Egyptian fraction decompositions of $1$ of length $3, \, 4, \, 5, \, 6, \,7, \, 8$ are $3,  \,  14,  \,  147,  \,  3462,  \,  294314,  \,  159330691$, respectively, see here  and OEIS A002966.
Note that your solution for $r=9$ corresponds to the length $10$ decomposition $$\frac{1}{4}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16} +\frac{1}{144}=1.$$
