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Following the presentation of the first Le Monde puzzle of the year, I tried a simulated annealing solution on an early morning in my hotel room. Here is the R code, which is unfortunately too rudimentary and too slow to be able to tackle n=1000.

#minimise sum_{i=1}^I x_i
#for 1le x_ile 2n+1, 1e ile I
# Ige n, x_i ne x_j
# a=x_i,b=x_j,i,jin I implies a+b=x_k for a kin I
n=6
m=2*n+1
complete=function(inde){
len=length(inde)
comp=outer(inde,inde,"+")
diag(comp)=inde
comp=sort(unique(comp[comp<m+1]))
while ((length(comp)>len)&&(length(comp)<m)){
inde=comp
len=length(inde)
comp=outer(inde,inde,"+")
diag(comp)=inde
comp=sort(unique(comp[comp<m+1]))
}
comp
}
move=function(inde,tempe){
ind=inde
# movin
if (length(ind)<m)
off=sample(ind,1,prob=ind)
inn=sample((1:m)[-ind],1)
newinde=sort(c(ind[ind!=off],inn))
newinde=complete(newinde)
if (tempe*log(runif(1))<(sum(ind)-sum(newinde)))
ind=newinde
}
# prunnin
if (length(ind)>n){
off=sample(ind,1,prob=ind)
newinde=sort(ind[ind!=off])
newinde=complete(newinde)
if (tempe*log(runif(1))<sum(ind)-sum(newinde))
ind=newinde
}
ind
}
T=10^4
fact=0.1
tpt=fact*seq(1,log(1+T),le=T)
inde=complete(sample(1:m,n))
recor=list(ind=inde,val=sum(inde))
for (t in 1:T){
inde=move(inde,tpt[t])
if (sum(inde)<recor$val){
recor=list(ind=inde,val=sum(inde))
}
}
print(recor)

The solution to the puzzle (given in the next Le Monde issue) is to take only the even digits, resulting in a minimum sum equal to n(n+1).