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The last and final Le Monde puzzle is a bit of a disappointment, to wit:

A 4×4 table is filled with positive and different integers. A 3×3 table is then deduced by adding four adjacent [i.e. sharing a common corner] entries of the original table. Similarly with a 2×2 table, summing up to a unique integer. What is the minimal value of this integer? And by how much does it increase if all 29 integers in the tables are different?

For the first question, the resulting integer writes down as the sum of the corner values, plus 3 times the sum of the side values, plus 9 times the sum of the 4 inner values [of the 4×4 table]. Hence, minimising the overall sum means taking the inner values as 1,2,3,4, the side values as 5,…,12, and the corner values as 13,…,16. Resulting in a total sum of 352. As checked in this computer code in APL by Jean-Louis:

This configuration does not produce 29 distinct values, but moving one value higher in one corner does: I experimented with different upper bounds on the numbers and 17 always provided with the smallest overall sum, 365.

firz=matrix(0,3,3)#second level
thirz=matrix(0,2,2)#third level
for (t in 1:1e8){
flor=matrix(sample(1:17,16),4,4)
for (i in 1:3) for (j in 1:3)
firz[i,j]=sum(flor[i:(i+1),j:(j+1)])
for (i in 1:2) for (j in 1:2)
thirz[i,j]=sum(firz[i:(i+1),j:(j+1)])
#last
if (length(unique(c(flor,firz,thirz)))==29)
solz=min(solz,sum(thirz))}

and a further simulated annealing attempt did not get me anywhere close to this solution.