According to the meaning of the question, there are * * * 2 (2n-1) × (2n-1) (n ∈ n and n≥2) diagonal lines in the square grid, that is, * *.

∴f(n)=(2n？ 1)2? (4n？ 3)4n？ 3,

∴f(n)=4n2？ 8n+44n？ 3= 14? 16n2？ 32n+ 164n？ 3= 14? (4n？ 3)2? 2(4n？ 3)+ 14n？ 3= 14? [(4n-3)+ 14n？ 3-2]

T= 14n？ 3,

∵n≥2, ∴ t ∈ (0, 15), y= 14(t+ 1t-2) in t ∈ (0, 15).

When t= 15, that is, n=2, there is a minimum value, and f (2) = 45.

So the answer is: 45.