The answer is yes. The largest such integer is 128. To see why, we first prove a lemma.

Lemma. Suppose n is a positive integer, and every integer from n+1 to 4n+35 (inclusive) can be written as a sum of distinct positive squares. Then every integer larger than n can be written as a sum of distinct positive squares.

Proof. Suppose not, and let k be the least counterexample. By hypothesis, k>=4n+36. Write k in the form k=4q+r, with 0<r<3, and note that n+9<=q<k. We now consider the four possible values for 4.

Suppose first that r=0. By the minimality of k, q can be written as a sum of distinct positive squares; say q=a²+b²+...+z². But then k=4q=(2a)²+(2b)²+...+(2z)², contradicting the assumption that k cannot be written as a sum of distinct positive squares. If r=1 then similar reasoning leads to k=4q+1=(2a)²+(2b)²+...+(2z)²+1, which is again a sum of distinct positive squares.

If r=2 then k=4q+2=4(q-2)+10. Since n+7<=q-2<k and k was chosen to be minimal, q-2 can be written as a sum of distinct positive squares; say q-2=a²+b²+...+z². Then

k=4(q-2)+10=(2a)²+(2b)²+...+(2z)²+1+9,

which is another sum of distinct positive squares.

Finally, if r=3 then k=4q+3=4(q-8)+1+9+25, and similar reasoning shows that this is a sum of distinct positive squares. This completes the proof of the lemma.

To complete the proof, simply write a computer program that verifies that 128 cannot be written as the sum of distinct squares and that all the integers from 128+1 to 4*128+35 (inclusive) can.

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