Agassi

Just by brute force a little python script gave two solutions $271389+32276=303665$ $183479+241186=424665$ of which the first would have a leading 0 in GARROS, which is ugly enough to opt for the second solution.

Here's the script (q&d):

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def val(a,b,c,d,e,f):
    return f+10*e+100*d+1000*c+10000*b+100000*a

from itertools import permutations 
def RolandGarrosAgassi():
    for perm in permutations(range(0, 10)):
        A=perm[0]
        D=perm[1]
        G=perm[2]
        I=perm[3]
        L=perm[4]
        N=perm[5]
        O=perm[6]
        R=perm[7]
        S=perm[8]
        Roland=val(R,O,L,A,N,D)
        Garros=val(G,A,R,R,O,S)
        Agassi=val(A,G,A,S,S,I)
        if Roland+Garros==Agassi:
            print ("{}+{}={}".format(Roland,Garros,Agassi))

RolandGarrosAgassi()

Analyse column by column, starting from the left. Since the sum is also a 6-digit number, $A$ must be greater than $R$ and $G$ . In the second column, $A$ is above $G$ , so $10$ from here is transferred to the first column, hence the equation $A = G + R + 1$ . In the third column, $A$ is below $L$ and $R$ , so it must be $A = L + R$ or $A = L + R + 1.$ But $A$ is already equal to $G + R + 1$ and $L$ and $G$ can not be equal, so $A = L + R$ and (no transfer from the 3rd to the 2nd column) $O + A = 10 + G$ . From these equations we can get $O + R = 9$ . Now things get a little complicated, and an attempt to obtain a $9 \times 9$ system of equations can be done by separation on different cases, when $10$ is transferred from a column to the one on the left or isn't. But, we know that $A$ cannot be $0,1,2,3$ and $9$ because $3 = G + R + 1$ implies $G = R$ , $A = 9$ implies $9 = L + R = O + R$ , so $O$ and $L$ would be equal, and so on. We restricted $A$ to be $4,5,6,7$ or $8$ , and by inserting these numbers for $A$ we can determine values of the remaining letters by trial and error; plugging $A = 4$ into summation yields $G = 2$ , $R = 1$ and so on ( $184479+241186=424665$ ), and, luckily enough, it is the solution (there is only one). There are many solutions that are "almost" correct and only one truly correct.