Romanian Master of Mathematics

Let $a={\min }_{i}a(i)$ and let $b={\min }_{i}b(i)$. Since a is not constant, there exists an $i$ such that $a=a(i)<a(i+1)$.

Claim 1. If $a=a(i)<a(i+1)$, then $b(i)=0$. Similarly, if $a=a(i)<a(i-1)$, then $b(i)=0$.

Proof. Otherwise the sum in $(\ast )$ contains a term $a(i+1)>a$ but no terms smaller than $a$, so the average is greater than $a$. $\square$

Say that $[i,j]$ is an a-segment if $a(i)=a(i+1)=\cdots =a(j)=0$ but $a(i-1)\ne 0\ne a(j+1)$; define a b-segment similarly. By Claim 1, the endpoints of any such segment satisfy $a(i)=b(i)=a(j)=b(j)=0$. Since the sequences are non-constant, each $i$ where $a(i)=0$ is contained in an a-segment.

Claim 2. Let $[i,j]$ be a b-segment, and let $k\in [i,j]$. Then $a(k)\le k-i$ (and, similarly, $a(k)\le j-k$).

Proof. Indeed, since $b(k)=0$, the elements of b with indices from $k-a(k)$ to $k+a(k)$ must all be zero as well. $\square$

We now show that every index is contained in either an a- or a b-segment. Since at least one index is contained in both, the conclusion follows.

Assume, to the contrary, that $a(i)$ and $b(i)$ are both positive for some index $i$; call such indices bad. Among all bad indices $i$, choose one maximising $\max (a(i),b(i))$; by symmetry, we may and will assume that this maximum is $a(i)$. We may and will also assume that either the index $i-1$ is not bad, or $a(i-1)<a(i)$ (otherwise change $i$ to $i-1$, repeat if necessary, recalling that a is not constant).

Consider the range of indices $\Delta =[i-b(i),i+b(i)]$, and the values a assumes at those indices. Some indices $j$ in $\Delta$ are bad; the corresponding values $a(j)$ do not exceed $a(i)$. Other indices $j$ in $\Delta$ are covered by several a- and b-segments. Each b-segment contributes at most $b(i)$ members nearest to one of its endpoints, so the average value of a over those indices does not exceed $(b(i)-1)/2<a(i)$ by Claim 2. The remaining indices $j$ in $\Delta$ all lie in a-segments, so the corresponding values $a(j)$ are all zero.

Combining all this, it follows that the average in the right-hand member of $(\ast )$ does not exceed $a(i)$. Moreover, if some a- or b-segment intersects $\Delta$, then the inequality is strict. Otherwise, $i-1$ is a bad index contained in $\Delta$, and $a(i-1)<a(i)$, so the inequality is again strict. This contradiction ends the proof and completes the solution.

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Alternative solution.

The solution has a few well-defined steps:

Lemma 1. Assume that $a(i)=M:=\max \mathbf{a}$; then $b(i+k)=0$ for all $k=-M,-M+1,\dots ,M$. In particular, $b(i-1)=b(i)=b(i+1)=0$, as $M\ge 1$.

Proof. Assume that $a(j)=a(j+1)=\cdots =a(j+s)=M>0$, and $a(j-1),a(j+s+1)<M$, where $i\in [j,j+s]$. Then $b(j)=0$, as otherwise $a(j)$ is the mean of at least three terms, all $\le M$, with at least one $<M$. For the same reason, $b(j+s)=0$ also. But then $b(j)$ is the mean of $2M+1$ terms of $b$, which must therefore all also be equal to 0. So $b(j+k)=0$ for all $k\in [-M,M]$. Iterating this argument gives $b(j+k)=0$ for all $k\in [-M,M+s]$, which implies the statement of the lemma. $\square$

Corollary. There exist $i$ such that $a(i)=0$ and $j$ such that $b(j)=0$.

Lemma 2. Suppose $\max a\ge \max b$. Generate $a^{\prime }$ by replacing all copies of $M=\max a$ with 1 in $a$. Then $a^{\prime }$ is $b$-harmonic, and $b$ is $a^{\prime }$-harmonic.

Proof. We start with another consequence of Lemma 1. Assume that $a(i)\ne M$; then none of the terms $a(i+k)$ with $k\in [-b(i),b(i)]$ equals $M$. Indeed, if $a(i+k)=M$ with $|k|\le b(i)\le M$, then by Lemma 1 we have $b(i)=b((i+k)-k)=0$, which yields $k=0$ and hence $a(i)=a(i+k)=M$. We can now check that the harmonic properties are preserved under replacing all copies of $M$ in $a$ with 1: If $a(i)\ne M$, then the harmonic property for $b(i)$ is unchanged. If $a(i)=M$, then $a^{\prime }(i)=1$ and $b(i-1)=b(i)=b(i+1)=0$, so $b(i)$ certainly has the $a^{\prime }(i)$-harmonic property; and If $a(i)=M$, then $b(i)=0$, and so $a^{\prime }(i)=1$ has the $b(i)$-harmonic property. If $a(i)\ne M$, then we have just shown that none of the terms in the statement of $a(i)$'s harmonic property are changed by this process, so it remains harmonic. $\square$

Lemma 3. We have $\min (a(i),b(i))=0$ for all $i$. Moreover, there exists an $i$ with $a(i)=b(i)=0$.

Proof. Both statements in the lemma are invariant under the procedure in Lemma 2. Apply this procedure repeatedly, to replace all instances of the maximum value in one of the sequences with 1, until both sequences consist of zeroes and ones. It suffices to check the lemma statement for the obtained pair of sequences. Suppose that $a(i)=b(i)=1$ for some $i$. Since the sequences remain non-constant, we may and will assume that $\min (a(i-1),b(i-1))=0$, say $a(i-1)=0$. But then the $b(i)$-harmonic property is violated for $a(i)$, as $a(i+1)\le 1$. Suppose now that there is no $i$ with $a(i)=b(i)=0$. This means that for every index $i$ we have either $a(i)=1$ and $b(i)=0$, or $a(i)=0$ and $b(i)=1$. There is a pair of adjacent indices having different types, so that $a(i)=b(i+1)=1$ and $a(i+1)=b(i)=0$. But then $b(i)$ violates the $a(i)$-harmonic property. $\square$

Lemma 3 readily yields that at least $N+1$ terms across both sequences are zeroes, as required.