The maximum number of fleas=20 There are totally 20 paths for the fleas(10 horizontal and 10 vertical) In each path there can be at most 1 flea or they will meet after at least 10 minutes. So the number of fleas<=20 Suppose,(i,j) denotes the cell at the meeting of i'th column and j'th row. At the initial position the fleas are in(1,1),(2,2),.........,(9,9) and they start to move in upwards direction. The flea in the (10,10) cell start to move downwards. The 9 more fleas are kept in (1,2),(2,3),(3,4),........(9,10) and they start to move to the right. Another flea is kept in (2,1) cell and it starts to move to left. In this case they will never meet.

I can only construct one with 20 as well, however i don't like RSM's argument "if two fleas are in the same path they will meet after at least 10 minutes" I don't agree for consider the following construction: The first two columns are completely filled with fleas (therefore 20 of them), now have them all move initially right, it is clear that they will never meet because as soon as the second column gets to the end, then the fleas swap positions with their neighbour and then the process reflects etc. The point is that this shows that you can have two fleas starting at the same path yet never meeting.

kamil9876 wrote: I can only construct one with 20 as well, however i don't like RSM's argument "if two fleas are in the same path they will meet after at least 10 minutes" I don't agree for consider the following construction: The first two columns are completely filled with fleas (therefore 20 of them), now have them all move initially right, it is clear that they will never meet because as soon as the second column gets to the end, then the fleas swap positions with their neighbour and then the process reflects etc. The point is that this shows that you can have two fleas starting at the same path yet never meeting. How can it be??? I said that no two fleas can be in the same path. Sitting at the same column when they move to right then the corresponding rows are their paths.

I am saying your proof is flawed because the statement "In each path there can be at most 1 flea or they will meet after at least 10 minutes." is incorrect. A counterexample to this statement is found in my model. To be explicity I will write it down, suppose that initially we have two Fleas A and B in the same row and they both move right: A B _ _ _ _ _ _ _ _ _ A B _ _ _ _ _ _ _ . . . . _ _ _ _ _ _ _ _ _ A B _ _ _ _ _ _ _ _ _ B A . . . etc. and the process repeats, A and B never meet yet they are on the same path.

36 is certainly possible ....<<.... ....<<.... ....<<.... ....<<.... vvvvv<^^^^ vvvv>^^^^^ ....>>.... ....>>.... ....>>.... ....>>.... so I'd guess the theoretical maximum 40 is possible but don't see an obvious construction.

Yes 40 is possible See in the attachment

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This solution provides some motivation for the construction of an example for $40.$ We claim that the answer is $\boxed{40}.$ Firstly, it is easily seen that more than $40$ fleas cannot be obtained because any row or column has at most two fleas. Let's now find a construction which attains $40$ fleas. First of all, observe that if flea $A$ is initially on a white square of the chessboard and flea $B$ is initially on a black square, then the fleas can never be on the same square. Hence, it suffices to construct an example where there are $20$ fleas on white squares, such that no two of them end up on the same square at any time. Indeed, an example of $20$ such fleas on black squares can be obtained analogously (or simply by reflecting over an axis of symmetry), and combining these two sets of $20$ fleas would finish the problem. So let's construct such an example of $20$ fleas on white squares. Notice that there must be exactly one flea in each row and column, as otherwise there are two in some row/column, and so because they're both on white squares they would coincide at some point. Label the rows $0, 1, 2, \cdots, 9$ from top to bottom and the columns $0, 1, 2, \cdots, 9$ from left to right. For each $0 \le i \le 9$, let $r_i$ denote the number of minutes after the fleas start jumping it takes for the flea in the $i$th row to reach the leftmost column (column $0$) Define $c_i$ similarly as the number of minutes it takes for the flea in the $i$th column to reach the uppermost row (row $0$). Working in modulo $18$ (since motion of the fleas is periodic with period $18$), we see that the condition of the flea in the $i$th row not intersecting the flea in the $j$th column is equivalent to: $$\{r_i \pm j\} \cap \{c_j \pm i\} = \emptyset,$$ where the sets are of residue classes modulo $18$. However, it can be seen that this is equivalent to: $$\{r_i \pm i\} \cap \{c_j \pm j\} = \emptyset,$$ where the sets are again residue classes modulo $18$. From here, we can focus on constructing the sets $\{r_i \pm i\}$ and $\{c_j \pm j\}$, and we arrive at the construction: $$(r_0, r_1, r_2, r_3, r_4, r_5, r_6, r_7, r_8, r_9) = (0, 1, 4, 3, 4, 13, 12, 13, 10, 9),$$ and $$(c_0, c_1, c_2, c_3, c_4, c_5, c_6, c_7, c_8, c_9) = (4, 11, 12, 7, 8, 17, 16, 3, 2, 13).$$ This yields the attached construction, and an analogous argument for fleas on black squares (or simply reflection in a vertical axis of symmetry) yields the desired $40$ fleas. $\square$

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Here's another construction for putting $20$ fleas on black squares (though the main idea is similar to the previous post): Let $\ell$ be the main diagonal consisting of black squares (note it doesn't matter even if take $\ell$ to the other diagonal). Claim: Two flees moving in perpendicular direction will meet iff if at some time, they are both simultaneously on $\ell$. Proof: Not hard. $\square$ Now note that modulo $18$, a flee will get at $\ell$ on at most two residues. Now consider the following construction. Note that the set of residues of the blues lines is $\{0,2,6,8\}$ and that of red lines is $\{ 4,-2,-4,-6 \}$. As these sets are disjoint, so we are done. $\blacksquare$