Let $\left(a_{1}, a_{2}, \ldots, a_{100}\right)$ be a permutation of $(1,2, \ldots, 100)$.

What is the maximum number of perfect squares that can occur among the numbers $a_{1}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \ldots, a_{1}+\ldots+a_{100} ?$
(Solved by Brieux Madeline-Derou)

The permutation

$$
(1,3,5, \ldots, 95,97,99,(100,98,6),(96,94,22),(92,90,38),(88,86,54)
$$

$(84,82,70),(80,78,76,10),(74,72,70,36),(68,66,64,62),(60,58,56,44,40,8,2)$,

$$
(52,50,48,46,42,26,12), 34,32,30,28,24,20,18,16,14,4)
$$

allows obtaining 60 perfect squares. Indeed, it is easy to verify that we get 50 perfect squares of the form $1+3+\ldots+(2 n+1)=(n+1)^{2}$, and 10 other squares by adding to $1+\ldots+99=50^{2}$ the terms in successive parentheses.

Suppose, for the sake of contradiction, that we have a permutation $\left(a_{1}, \ldots, a_{100}\right)$ allowing us to obtain 61 perfect squares. We start by noting that $1+2+\ldots+100=5050$, so the maximum square achievable is $71^{2}=5041$.

It is clear that if one of the sums $a_{1}+\ldots+a_{i}$ is a square, the next square we obtain is strictly greater than this one, since all the $a_{i}$ are strictly positive.

However, to go from $a^{2}$ to $(a+1)^{2}$, one must add $2 a+1$, and thus at least one odd number must be summed. Since there are only 50 odd numbers between 1 and 100, we can only go from $a^{2}$ to $(a+1)^{2}$ at most 50 times.

Thus, the other "transitions" between two squares require going from $a^{2}$ to $(a+k)^{2}$ with $k \geq 2$. Therefore, the 61st square would be at least $50+2 \cdot(61-50)=72$. But $72^{2}>5050$, which is absurd.

The number sought is therefore 60.